What happens when creating a variable? [closed]

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void Method1()

string str =
client.GetString("http://msdn.microsoft.com");




What exactly happens when the first line in Method1 is carried out?



I understand memory is set aside for the string variable str, but does the right-hand side of the statement also get carried out at this stage? i.e. does it actually retrieve a value for the right-hand side?










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closed as too broad by Steve, D-Shih, techraf, Nick A, Servy Aug 13 at 15:49


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 2




    @Alejandro: Right hand side of the assignment.
    – Amadan
    Aug 13 at 12:25






  • 4




    If the RHS didn't get carried out, what would be the point of having it?
    – Scott Hunter
    Aug 13 at 12:25






  • 3




    Yes. This is declaration (string str) and initialisation (str = ...) in one statement.
    – Amadan
    Aug 13 at 12:26






  • 2




    Code isn’t run line by line, especially when lines are cut in the middle of an actual code unit. Therefore it’s hard to say what you want to know exactly here and what to answer. There is only one line, which calls a method and stores the result in a variable. Compiler may throw the whole variable assignment out since it’s not used anywhere so there might not be any memory allocations.
    – Sami Kuhmonen
    Aug 13 at 12:27






  • 2




    "whether it was worthwhile storing a variable and then doing work on that variable later on" - quite possibly; depends entirely on context, though
    – Marc Gravell♦
    Aug 13 at 12:46














up vote
9
down vote

favorite
1












void Method1()

string str =
client.GetString("http://msdn.microsoft.com");




What exactly happens when the first line in Method1 is carried out?



I understand memory is set aside for the string variable str, but does the right-hand side of the statement also get carried out at this stage? i.e. does it actually retrieve a value for the right-hand side?










share|improve this question















closed as too broad by Steve, D-Shih, techraf, Nick A, Servy Aug 13 at 15:49


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 2




    @Alejandro: Right hand side of the assignment.
    – Amadan
    Aug 13 at 12:25






  • 4




    If the RHS didn't get carried out, what would be the point of having it?
    – Scott Hunter
    Aug 13 at 12:25






  • 3




    Yes. This is declaration (string str) and initialisation (str = ...) in one statement.
    – Amadan
    Aug 13 at 12:26






  • 2




    Code isn’t run line by line, especially when lines are cut in the middle of an actual code unit. Therefore it’s hard to say what you want to know exactly here and what to answer. There is only one line, which calls a method and stores the result in a variable. Compiler may throw the whole variable assignment out since it’s not used anywhere so there might not be any memory allocations.
    – Sami Kuhmonen
    Aug 13 at 12:27






  • 2




    "whether it was worthwhile storing a variable and then doing work on that variable later on" - quite possibly; depends entirely on context, though
    – Marc Gravell♦
    Aug 13 at 12:46












up vote
9
down vote

favorite
1









up vote
9
down vote

favorite
1






1





void Method1()

string str =
client.GetString("http://msdn.microsoft.com");




What exactly happens when the first line in Method1 is carried out?



I understand memory is set aside for the string variable str, but does the right-hand side of the statement also get carried out at this stage? i.e. does it actually retrieve a value for the right-hand side?










share|improve this question















void Method1()

string str =
client.GetString("http://msdn.microsoft.com");




What exactly happens when the first line in Method1 is carried out?



I understand memory is set aside for the string variable str, but does the right-hand side of the statement also get carried out at this stage? i.e. does it actually retrieve a value for the right-hand side?







c# variables






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Aug 13 at 15:23









TylerH

15k104967




15k104967










asked Aug 13 at 12:23









jim bob

565




565




closed as too broad by Steve, D-Shih, techraf, Nick A, Servy Aug 13 at 15:49


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as too broad by Steve, D-Shih, techraf, Nick A, Servy Aug 13 at 15:49


Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    @Alejandro: Right hand side of the assignment.
    – Amadan
    Aug 13 at 12:25






  • 4




    If the RHS didn't get carried out, what would be the point of having it?
    – Scott Hunter
    Aug 13 at 12:25






  • 3




    Yes. This is declaration (string str) and initialisation (str = ...) in one statement.
    – Amadan
    Aug 13 at 12:26






  • 2




    Code isn’t run line by line, especially when lines are cut in the middle of an actual code unit. Therefore it’s hard to say what you want to know exactly here and what to answer. There is only one line, which calls a method and stores the result in a variable. Compiler may throw the whole variable assignment out since it’s not used anywhere so there might not be any memory allocations.
    – Sami Kuhmonen
    Aug 13 at 12:27






  • 2




    "whether it was worthwhile storing a variable and then doing work on that variable later on" - quite possibly; depends entirely on context, though
    – Marc Gravell♦
    Aug 13 at 12:46












  • 2




    @Alejandro: Right hand side of the assignment.
    – Amadan
    Aug 13 at 12:25






  • 4




    If the RHS didn't get carried out, what would be the point of having it?
    – Scott Hunter
    Aug 13 at 12:25






  • 3




    Yes. This is declaration (string str) and initialisation (str = ...) in one statement.
    – Amadan
    Aug 13 at 12:26






  • 2




    Code isn’t run line by line, especially when lines are cut in the middle of an actual code unit. Therefore it’s hard to say what you want to know exactly here and what to answer. There is only one line, which calls a method and stores the result in a variable. Compiler may throw the whole variable assignment out since it’s not used anywhere so there might not be any memory allocations.
    – Sami Kuhmonen
    Aug 13 at 12:27






  • 2




    "whether it was worthwhile storing a variable and then doing work on that variable later on" - quite possibly; depends entirely on context, though
    – Marc Gravell♦
    Aug 13 at 12:46







2




2




@Alejandro: Right hand side of the assignment.
– Amadan
Aug 13 at 12:25




@Alejandro: Right hand side of the assignment.
– Amadan
Aug 13 at 12:25




4




4




If the RHS didn't get carried out, what would be the point of having it?
– Scott Hunter
Aug 13 at 12:25




If the RHS didn't get carried out, what would be the point of having it?
– Scott Hunter
Aug 13 at 12:25




3




3




Yes. This is declaration (string str) and initialisation (str = ...) in one statement.
– Amadan
Aug 13 at 12:26




Yes. This is declaration (string str) and initialisation (str = ...) in one statement.
– Amadan
Aug 13 at 12:26




2




2




Code isn’t run line by line, especially when lines are cut in the middle of an actual code unit. Therefore it’s hard to say what you want to know exactly here and what to answer. There is only one line, which calls a method and stores the result in a variable. Compiler may throw the whole variable assignment out since it’s not used anywhere so there might not be any memory allocations.
– Sami Kuhmonen
Aug 13 at 12:27




Code isn’t run line by line, especially when lines are cut in the middle of an actual code unit. Therefore it’s hard to say what you want to know exactly here and what to answer. There is only one line, which calls a method and stores the result in a variable. Compiler may throw the whole variable assignment out since it’s not used anywhere so there might not be any memory allocations.
– Sami Kuhmonen
Aug 13 at 12:27




2




2




"whether it was worthwhile storing a variable and then doing work on that variable later on" - quite possibly; depends entirely on context, though
– Marc Gravell♦
Aug 13 at 12:46




"whether it was worthwhile storing a variable and then doing work on that variable later on" - quite possibly; depends entirely on context, though
– Marc Gravell♦
Aug 13 at 12:46












3 Answers
3






active

oldest

votes

















up vote
13
down vote













This depends a lot on what you do next. There's a very good chance that the compiler will actually remove str completely if you don't use it (unless you use it in the next step, or the things you do between now and then are "net zero" in terms of stack positions). It will still execute the call to client.GetString(...), of course; the question is what does it do with the result? There's various ways the compiler can interpret this:



  • as a local:

stack space for the local is reserved as part of the stackframe entry; after the call to GetString the compiler emits stloc (or a variant)



  • as an ambient stack value

no explicit stack space is reserved for the local; after the GetString() it is simply left where it is for the next operation to consume (for example, this would be perfect if followed by a static call like Console.WriteLine(str);); it might also be cloned (dup) if needed multiple times



  • popped

no explicit stack space is reserved for the local; after the GetString() it is simply dropped (pop)



  • as a field

this would apply for iterator blocks and async methods; very complicated to explain



Ultimately, if you really want to know, you need to look at the real code, then look at the IL - ideally compiled in "release" mode.



You can see examples of some of these in this test code on sharplab.io



or copied here:



void Method1_Popped()

string str = client.GetString("http://msdn.microsoft.com");

void Method2_LeftOnStack()

string str = client.GetString("http://msdn.microsoft.com");
Console.WriteLine(str);

void Method3_Local()

string str = client.GetString("http://msdn.microsoft.com");
for(int i = 0;i < 3 ; i++) DoSomethingElse();
Console.WriteLine(str);



becomes:



.method private hidebysig 
instance void Method1_Popped () cil managed

// Method begins at RVA 0x2050
// Code size 18 (0x12)
.maxstack 8

IL_0000: ldarg.0
IL_0001: ldfld class SomeClient Foo::client
IL_0006: ldstr "http://msdn.microsoft.com"
IL_000b: callvirt instance string SomeClient::GetString(string)
IL_0010: pop
IL_0011: ret
// end of method Foo::Method1_Popped

.method private hidebysig
instance void Method2_LeftOnStack () cil managed

// Method begins at RVA 0x2063
// Code size 22 (0x16)
.maxstack 8

IL_0000: ldarg.0
IL_0001: ldfld class SomeClient Foo::client
IL_0006: ldstr "http://msdn.microsoft.com"
IL_000b: callvirt instance string SomeClient::GetString(string)
IL_0010: call void [mscorlib]System.Console::WriteLine(string)
IL_0015: ret
// end of method Foo::Method2_LeftOnStack

.method private hidebysig
instance void Method3_Local () cil managed

// Method begins at RVA 0x207c
// Code size 42 (0x2a)
.maxstack 2
.locals init (
[0] string,
[1] int32
)

IL_0000: ldarg.0
IL_0001: ldfld class SomeClient Foo::client
IL_0006: ldstr "http://msdn.microsoft.com"
IL_000b: callvirt instance string SomeClient::GetString(string)
IL_0010: stloc.0
IL_0011: ldc.i4.0
IL_0012: stloc.1
// sequence point: hidden
IL_0013: br.s IL_001f
// loop start (head: IL_001f)
IL_0015: ldarg.0
IL_0016: call instance void Foo::DoSomethingElse()
IL_001b: ldloc.1
IL_001c: ldc.i4.1
IL_001d: add
IL_001e: stloc.1

IL_001f: ldloc.1
IL_0020: ldc.i4.3
IL_0021: blt.s IL_0015
// end loop

IL_0023: ldloc.0
IL_0024: call void [mscorlib]System.Console::WriteLine(string)
IL_0029: ret
// end of method Foo::Method3_Local


or as ASM:



Foo.Method1_Popped()
L0000: mov ecx, [ecx+0x4]
L0003: mov edx, [0xe42586c]
L0009: cmp [ecx], ecx
L000b: call dword [0x2ef71758]
L0011: ret

Foo.Method2_LeftOnStack()
L0000: push ebp
L0001: mov ebp, esp
L0003: mov ecx, [ecx+0x4]
L0006: mov edx, [0xe42586c]
L000c: cmp [ecx], ecx
L000e: call dword [0x2ef71758]
L0014: mov ecx, eax
L0016: call System.Console.WriteLine(System.String)
L001b: pop ebp
L001c: ret

Foo.Method3_Local()
L0000: push ebp
L0001: mov ebp, esp
L0003: mov ecx, [ecx+0x4]
L0006: mov edx, [0xe42586c]
L000c: cmp [ecx], ecx
L000e: call dword [0x2ef71758]
L0014: mov ecx, eax
L0016: call System.Console.WriteLine(System.String)
L001b: pop ebp
L001c: ret





share|improve this answer






















  • sorry i have no idea what you said, but i will save the reply and hopefully one day i can come back to it and be able to understand it :) thanks marc, i've also saved the new edits
    – jim bob
    Aug 13 at 12:37











  • @jimbob I've added more context that may help you... or not :) and note: this is just in terms of IL; the JIT compiler can then do even more voodoo; there is also a "JIT Asm" tab on sharplab.io
    – Marc Gravell♦
    Aug 13 at 12:37











  • @jimbob ultimately, there's no way of answering this that doesn't involve talking about stackframes and IL...
    – Marc Gravell♦
    Aug 13 at 12:40










  • It seems you made a copy&paste mistake in the assembly listing: Foo.Method3_Local is identical to Foo.Method2_LeftOnStack.
    – Marc Schütz
    Aug 13 at 16:13

















up vote
1
down vote













Memory allocation depends on how exactly the variable is used and where it's declared. In this case, being a local variable within a method*, memory is reserved as soon as the method is called (and not when execution reaches the place of its declaration), regardless of what could happen afterwards. So memory is set aside even if the client.GetString("http://msdn.microsoft.com") isn't called at all (couldn't happen at all here, but in more complex code that's possible).



Note that you mention




when the first line in Method1 is carried out




This method has a single line, which consist in declaring a variable and assigning a value to it from calling another method. The fact that you've written it as two physical lines is irrelevant, as logically, your whole code consist of a single step. Again, variable "declaration" and allocation happens as soon as the method is called, and the rest of the line happens when execution reaches that point.



The execution of the line actually has two stages: the first, the GetString method gets called. Second, its return value gets assigned to the local variable.



As Marc Gravell pointed out, things can become more complex. The compiler may decide to simply not create the variable at all, or arrange things differently, as long as that produces the same result (something known as compiler optimizations). The rest of this answer assumes that the compiler didn't optimized anything and created a binary that matches exactly the given code, but in release builds we can expect some differences.



*(and not captured by a lambda)






share|improve this answer






















  • note - it is very possible (likely, even), that no stack space is actually reserved here; many local variables are elided completely by the compiler
    – Marc Gravell♦
    Aug 13 at 12:50










  • @MarcGravell Sure, in the case the compiler optimizes out the variable (probably in this case it does). For the simplicity sake, I'm assuming that the compiler compiles the code exactly as stands in the question, no optimizations involved. In real life, the actual binary might differ significantly from the code, as long as it produces the same result.
    – Alejandro
    Aug 13 at 13:06






  • 1




    One possible source of confusion: How can the compiler reserve memory at the start of the method when it doesn't even know how long the string is going to be? It might make sense to point out that space for the string content gets allocated somewhere within the GetString() call, and Method1 just stores a reference (to avoid the ugly word "pointer") to that space, with the amount of storage needed for the reference being always the same, independent of string length.
    – Guntram Blohm
    Aug 13 at 15:34


















up vote
0
down vote













Yes, in this case, the assignment happens immediately. Later on you'll learn about funny things like lazy loading, async and delegates and then you'll see instances where variables aren't initialised immediately. But don't worry, by the time you learn about those things, you'll know enough that it won't be difficult to understand.



And congrats on learning the most beautiful programming language in the world.






share|improve this answer



























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    13
    down vote













    This depends a lot on what you do next. There's a very good chance that the compiler will actually remove str completely if you don't use it (unless you use it in the next step, or the things you do between now and then are "net zero" in terms of stack positions). It will still execute the call to client.GetString(...), of course; the question is what does it do with the result? There's various ways the compiler can interpret this:



    • as a local:

    stack space for the local is reserved as part of the stackframe entry; after the call to GetString the compiler emits stloc (or a variant)



    • as an ambient stack value

    no explicit stack space is reserved for the local; after the GetString() it is simply left where it is for the next operation to consume (for example, this would be perfect if followed by a static call like Console.WriteLine(str);); it might also be cloned (dup) if needed multiple times



    • popped

    no explicit stack space is reserved for the local; after the GetString() it is simply dropped (pop)



    • as a field

    this would apply for iterator blocks and async methods; very complicated to explain



    Ultimately, if you really want to know, you need to look at the real code, then look at the IL - ideally compiled in "release" mode.



    You can see examples of some of these in this test code on sharplab.io



    or copied here:



    void Method1_Popped()

    string str = client.GetString("http://msdn.microsoft.com");

    void Method2_LeftOnStack()

    string str = client.GetString("http://msdn.microsoft.com");
    Console.WriteLine(str);

    void Method3_Local()

    string str = client.GetString("http://msdn.microsoft.com");
    for(int i = 0;i < 3 ; i++) DoSomethingElse();
    Console.WriteLine(str);



    becomes:



    .method private hidebysig 
    instance void Method1_Popped () cil managed

    // Method begins at RVA 0x2050
    // Code size 18 (0x12)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: pop
    IL_0011: ret
    // end of method Foo::Method1_Popped

    .method private hidebysig
    instance void Method2_LeftOnStack () cil managed

    // Method begins at RVA 0x2063
    // Code size 22 (0x16)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: call void [mscorlib]System.Console::WriteLine(string)
    IL_0015: ret
    // end of method Foo::Method2_LeftOnStack

    .method private hidebysig
    instance void Method3_Local () cil managed

    // Method begins at RVA 0x207c
    // Code size 42 (0x2a)
    .maxstack 2
    .locals init (
    [0] string,
    [1] int32
    )

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: stloc.0
    IL_0011: ldc.i4.0
    IL_0012: stloc.1
    // sequence point: hidden
    IL_0013: br.s IL_001f
    // loop start (head: IL_001f)
    IL_0015: ldarg.0
    IL_0016: call instance void Foo::DoSomethingElse()
    IL_001b: ldloc.1
    IL_001c: ldc.i4.1
    IL_001d: add
    IL_001e: stloc.1

    IL_001f: ldloc.1
    IL_0020: ldc.i4.3
    IL_0021: blt.s IL_0015
    // end loop

    IL_0023: ldloc.0
    IL_0024: call void [mscorlib]System.Console::WriteLine(string)
    IL_0029: ret
    // end of method Foo::Method3_Local


    or as ASM:



    Foo.Method1_Popped()
    L0000: mov ecx, [ecx+0x4]
    L0003: mov edx, [0xe42586c]
    L0009: cmp [ecx], ecx
    L000b: call dword [0x2ef71758]
    L0011: ret

    Foo.Method2_LeftOnStack()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret

    Foo.Method3_Local()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret





    share|improve this answer






















    • sorry i have no idea what you said, but i will save the reply and hopefully one day i can come back to it and be able to understand it :) thanks marc, i've also saved the new edits
      – jim bob
      Aug 13 at 12:37











    • @jimbob I've added more context that may help you... or not :) and note: this is just in terms of IL; the JIT compiler can then do even more voodoo; there is also a "JIT Asm" tab on sharplab.io
      – Marc Gravell♦
      Aug 13 at 12:37











    • @jimbob ultimately, there's no way of answering this that doesn't involve talking about stackframes and IL...
      – Marc Gravell♦
      Aug 13 at 12:40










    • It seems you made a copy&paste mistake in the assembly listing: Foo.Method3_Local is identical to Foo.Method2_LeftOnStack.
      – Marc Schütz
      Aug 13 at 16:13














    up vote
    13
    down vote













    This depends a lot on what you do next. There's a very good chance that the compiler will actually remove str completely if you don't use it (unless you use it in the next step, or the things you do between now and then are "net zero" in terms of stack positions). It will still execute the call to client.GetString(...), of course; the question is what does it do with the result? There's various ways the compiler can interpret this:



    • as a local:

    stack space for the local is reserved as part of the stackframe entry; after the call to GetString the compiler emits stloc (or a variant)



    • as an ambient stack value

    no explicit stack space is reserved for the local; after the GetString() it is simply left where it is for the next operation to consume (for example, this would be perfect if followed by a static call like Console.WriteLine(str);); it might also be cloned (dup) if needed multiple times



    • popped

    no explicit stack space is reserved for the local; after the GetString() it is simply dropped (pop)



    • as a field

    this would apply for iterator blocks and async methods; very complicated to explain



    Ultimately, if you really want to know, you need to look at the real code, then look at the IL - ideally compiled in "release" mode.



    You can see examples of some of these in this test code on sharplab.io



    or copied here:



    void Method1_Popped()

    string str = client.GetString("http://msdn.microsoft.com");

    void Method2_LeftOnStack()

    string str = client.GetString("http://msdn.microsoft.com");
    Console.WriteLine(str);

    void Method3_Local()

    string str = client.GetString("http://msdn.microsoft.com");
    for(int i = 0;i < 3 ; i++) DoSomethingElse();
    Console.WriteLine(str);



    becomes:



    .method private hidebysig 
    instance void Method1_Popped () cil managed

    // Method begins at RVA 0x2050
    // Code size 18 (0x12)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: pop
    IL_0011: ret
    // end of method Foo::Method1_Popped

    .method private hidebysig
    instance void Method2_LeftOnStack () cil managed

    // Method begins at RVA 0x2063
    // Code size 22 (0x16)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: call void [mscorlib]System.Console::WriteLine(string)
    IL_0015: ret
    // end of method Foo::Method2_LeftOnStack

    .method private hidebysig
    instance void Method3_Local () cil managed

    // Method begins at RVA 0x207c
    // Code size 42 (0x2a)
    .maxstack 2
    .locals init (
    [0] string,
    [1] int32
    )

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: stloc.0
    IL_0011: ldc.i4.0
    IL_0012: stloc.1
    // sequence point: hidden
    IL_0013: br.s IL_001f
    // loop start (head: IL_001f)
    IL_0015: ldarg.0
    IL_0016: call instance void Foo::DoSomethingElse()
    IL_001b: ldloc.1
    IL_001c: ldc.i4.1
    IL_001d: add
    IL_001e: stloc.1

    IL_001f: ldloc.1
    IL_0020: ldc.i4.3
    IL_0021: blt.s IL_0015
    // end loop

    IL_0023: ldloc.0
    IL_0024: call void [mscorlib]System.Console::WriteLine(string)
    IL_0029: ret
    // end of method Foo::Method3_Local


    or as ASM:



    Foo.Method1_Popped()
    L0000: mov ecx, [ecx+0x4]
    L0003: mov edx, [0xe42586c]
    L0009: cmp [ecx], ecx
    L000b: call dword [0x2ef71758]
    L0011: ret

    Foo.Method2_LeftOnStack()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret

    Foo.Method3_Local()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret





    share|improve this answer






















    • sorry i have no idea what you said, but i will save the reply and hopefully one day i can come back to it and be able to understand it :) thanks marc, i've also saved the new edits
      – jim bob
      Aug 13 at 12:37











    • @jimbob I've added more context that may help you... or not :) and note: this is just in terms of IL; the JIT compiler can then do even more voodoo; there is also a "JIT Asm" tab on sharplab.io
      – Marc Gravell♦
      Aug 13 at 12:37











    • @jimbob ultimately, there's no way of answering this that doesn't involve talking about stackframes and IL...
      – Marc Gravell♦
      Aug 13 at 12:40










    • It seems you made a copy&paste mistake in the assembly listing: Foo.Method3_Local is identical to Foo.Method2_LeftOnStack.
      – Marc Schütz
      Aug 13 at 16:13












    up vote
    13
    down vote










    up vote
    13
    down vote









    This depends a lot on what you do next. There's a very good chance that the compiler will actually remove str completely if you don't use it (unless you use it in the next step, or the things you do between now and then are "net zero" in terms of stack positions). It will still execute the call to client.GetString(...), of course; the question is what does it do with the result? There's various ways the compiler can interpret this:



    • as a local:

    stack space for the local is reserved as part of the stackframe entry; after the call to GetString the compiler emits stloc (or a variant)



    • as an ambient stack value

    no explicit stack space is reserved for the local; after the GetString() it is simply left where it is for the next operation to consume (for example, this would be perfect if followed by a static call like Console.WriteLine(str);); it might also be cloned (dup) if needed multiple times



    • popped

    no explicit stack space is reserved for the local; after the GetString() it is simply dropped (pop)



    • as a field

    this would apply for iterator blocks and async methods; very complicated to explain



    Ultimately, if you really want to know, you need to look at the real code, then look at the IL - ideally compiled in "release" mode.



    You can see examples of some of these in this test code on sharplab.io



    or copied here:



    void Method1_Popped()

    string str = client.GetString("http://msdn.microsoft.com");

    void Method2_LeftOnStack()

    string str = client.GetString("http://msdn.microsoft.com");
    Console.WriteLine(str);

    void Method3_Local()

    string str = client.GetString("http://msdn.microsoft.com");
    for(int i = 0;i < 3 ; i++) DoSomethingElse();
    Console.WriteLine(str);



    becomes:



    .method private hidebysig 
    instance void Method1_Popped () cil managed

    // Method begins at RVA 0x2050
    // Code size 18 (0x12)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: pop
    IL_0011: ret
    // end of method Foo::Method1_Popped

    .method private hidebysig
    instance void Method2_LeftOnStack () cil managed

    // Method begins at RVA 0x2063
    // Code size 22 (0x16)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: call void [mscorlib]System.Console::WriteLine(string)
    IL_0015: ret
    // end of method Foo::Method2_LeftOnStack

    .method private hidebysig
    instance void Method3_Local () cil managed

    // Method begins at RVA 0x207c
    // Code size 42 (0x2a)
    .maxstack 2
    .locals init (
    [0] string,
    [1] int32
    )

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: stloc.0
    IL_0011: ldc.i4.0
    IL_0012: stloc.1
    // sequence point: hidden
    IL_0013: br.s IL_001f
    // loop start (head: IL_001f)
    IL_0015: ldarg.0
    IL_0016: call instance void Foo::DoSomethingElse()
    IL_001b: ldloc.1
    IL_001c: ldc.i4.1
    IL_001d: add
    IL_001e: stloc.1

    IL_001f: ldloc.1
    IL_0020: ldc.i4.3
    IL_0021: blt.s IL_0015
    // end loop

    IL_0023: ldloc.0
    IL_0024: call void [mscorlib]System.Console::WriteLine(string)
    IL_0029: ret
    // end of method Foo::Method3_Local


    or as ASM:



    Foo.Method1_Popped()
    L0000: mov ecx, [ecx+0x4]
    L0003: mov edx, [0xe42586c]
    L0009: cmp [ecx], ecx
    L000b: call dword [0x2ef71758]
    L0011: ret

    Foo.Method2_LeftOnStack()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret

    Foo.Method3_Local()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret





    share|improve this answer














    This depends a lot on what you do next. There's a very good chance that the compiler will actually remove str completely if you don't use it (unless you use it in the next step, or the things you do between now and then are "net zero" in terms of stack positions). It will still execute the call to client.GetString(...), of course; the question is what does it do with the result? There's various ways the compiler can interpret this:



    • as a local:

    stack space for the local is reserved as part of the stackframe entry; after the call to GetString the compiler emits stloc (or a variant)



    • as an ambient stack value

    no explicit stack space is reserved for the local; after the GetString() it is simply left where it is for the next operation to consume (for example, this would be perfect if followed by a static call like Console.WriteLine(str);); it might also be cloned (dup) if needed multiple times



    • popped

    no explicit stack space is reserved for the local; after the GetString() it is simply dropped (pop)



    • as a field

    this would apply for iterator blocks and async methods; very complicated to explain



    Ultimately, if you really want to know, you need to look at the real code, then look at the IL - ideally compiled in "release" mode.



    You can see examples of some of these in this test code on sharplab.io



    or copied here:



    void Method1_Popped()

    string str = client.GetString("http://msdn.microsoft.com");

    void Method2_LeftOnStack()

    string str = client.GetString("http://msdn.microsoft.com");
    Console.WriteLine(str);

    void Method3_Local()

    string str = client.GetString("http://msdn.microsoft.com");
    for(int i = 0;i < 3 ; i++) DoSomethingElse();
    Console.WriteLine(str);



    becomes:



    .method private hidebysig 
    instance void Method1_Popped () cil managed

    // Method begins at RVA 0x2050
    // Code size 18 (0x12)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: pop
    IL_0011: ret
    // end of method Foo::Method1_Popped

    .method private hidebysig
    instance void Method2_LeftOnStack () cil managed

    // Method begins at RVA 0x2063
    // Code size 22 (0x16)
    .maxstack 8

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: call void [mscorlib]System.Console::WriteLine(string)
    IL_0015: ret
    // end of method Foo::Method2_LeftOnStack

    .method private hidebysig
    instance void Method3_Local () cil managed

    // Method begins at RVA 0x207c
    // Code size 42 (0x2a)
    .maxstack 2
    .locals init (
    [0] string,
    [1] int32
    )

    IL_0000: ldarg.0
    IL_0001: ldfld class SomeClient Foo::client
    IL_0006: ldstr "http://msdn.microsoft.com"
    IL_000b: callvirt instance string SomeClient::GetString(string)
    IL_0010: stloc.0
    IL_0011: ldc.i4.0
    IL_0012: stloc.1
    // sequence point: hidden
    IL_0013: br.s IL_001f
    // loop start (head: IL_001f)
    IL_0015: ldarg.0
    IL_0016: call instance void Foo::DoSomethingElse()
    IL_001b: ldloc.1
    IL_001c: ldc.i4.1
    IL_001d: add
    IL_001e: stloc.1

    IL_001f: ldloc.1
    IL_0020: ldc.i4.3
    IL_0021: blt.s IL_0015
    // end loop

    IL_0023: ldloc.0
    IL_0024: call void [mscorlib]System.Console::WriteLine(string)
    IL_0029: ret
    // end of method Foo::Method3_Local


    or as ASM:



    Foo.Method1_Popped()
    L0000: mov ecx, [ecx+0x4]
    L0003: mov edx, [0xe42586c]
    L0009: cmp [ecx], ecx
    L000b: call dword [0x2ef71758]
    L0011: ret

    Foo.Method2_LeftOnStack()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret

    Foo.Method3_Local()
    L0000: push ebp
    L0001: mov ebp, esp
    L0003: mov ecx, [ecx+0x4]
    L0006: mov edx, [0xe42586c]
    L000c: cmp [ecx], ecx
    L000e: call dword [0x2ef71758]
    L0014: mov ecx, eax
    L0016: call System.Console.WriteLine(System.String)
    L001b: pop ebp
    L001c: ret






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Aug 13 at 12:45

























    answered Aug 13 at 12:32









    Marc Gravell♦

    758k18920892512




    758k18920892512











    • sorry i have no idea what you said, but i will save the reply and hopefully one day i can come back to it and be able to understand it :) thanks marc, i've also saved the new edits
      – jim bob
      Aug 13 at 12:37











    • @jimbob I've added more context that may help you... or not :) and note: this is just in terms of IL; the JIT compiler can then do even more voodoo; there is also a "JIT Asm" tab on sharplab.io
      – Marc Gravell♦
      Aug 13 at 12:37











    • @jimbob ultimately, there's no way of answering this that doesn't involve talking about stackframes and IL...
      – Marc Gravell♦
      Aug 13 at 12:40










    • It seems you made a copy&paste mistake in the assembly listing: Foo.Method3_Local is identical to Foo.Method2_LeftOnStack.
      – Marc Schütz
      Aug 13 at 16:13
















    • sorry i have no idea what you said, but i will save the reply and hopefully one day i can come back to it and be able to understand it :) thanks marc, i've also saved the new edits
      – jim bob
      Aug 13 at 12:37











    • @jimbob I've added more context that may help you... or not :) and note: this is just in terms of IL; the JIT compiler can then do even more voodoo; there is also a "JIT Asm" tab on sharplab.io
      – Marc Gravell♦
      Aug 13 at 12:37











    • @jimbob ultimately, there's no way of answering this that doesn't involve talking about stackframes and IL...
      – Marc Gravell♦
      Aug 13 at 12:40










    • It seems you made a copy&paste mistake in the assembly listing: Foo.Method3_Local is identical to Foo.Method2_LeftOnStack.
      – Marc Schütz
      Aug 13 at 16:13















    sorry i have no idea what you said, but i will save the reply and hopefully one day i can come back to it and be able to understand it :) thanks marc, i've also saved the new edits
    – jim bob
    Aug 13 at 12:37





    sorry i have no idea what you said, but i will save the reply and hopefully one day i can come back to it and be able to understand it :) thanks marc, i've also saved the new edits
    – jim bob
    Aug 13 at 12:37













    @jimbob I've added more context that may help you... or not :) and note: this is just in terms of IL; the JIT compiler can then do even more voodoo; there is also a "JIT Asm" tab on sharplab.io
    – Marc Gravell♦
    Aug 13 at 12:37





    @jimbob I've added more context that may help you... or not :) and note: this is just in terms of IL; the JIT compiler can then do even more voodoo; there is also a "JIT Asm" tab on sharplab.io
    – Marc Gravell♦
    Aug 13 at 12:37













    @jimbob ultimately, there's no way of answering this that doesn't involve talking about stackframes and IL...
    – Marc Gravell♦
    Aug 13 at 12:40




    @jimbob ultimately, there's no way of answering this that doesn't involve talking about stackframes and IL...
    – Marc Gravell♦
    Aug 13 at 12:40












    It seems you made a copy&paste mistake in the assembly listing: Foo.Method3_Local is identical to Foo.Method2_LeftOnStack.
    – Marc Schütz
    Aug 13 at 16:13




    It seems you made a copy&paste mistake in the assembly listing: Foo.Method3_Local is identical to Foo.Method2_LeftOnStack.
    – Marc Schütz
    Aug 13 at 16:13












    up vote
    1
    down vote













    Memory allocation depends on how exactly the variable is used and where it's declared. In this case, being a local variable within a method*, memory is reserved as soon as the method is called (and not when execution reaches the place of its declaration), regardless of what could happen afterwards. So memory is set aside even if the client.GetString("http://msdn.microsoft.com") isn't called at all (couldn't happen at all here, but in more complex code that's possible).



    Note that you mention




    when the first line in Method1 is carried out




    This method has a single line, which consist in declaring a variable and assigning a value to it from calling another method. The fact that you've written it as two physical lines is irrelevant, as logically, your whole code consist of a single step. Again, variable "declaration" and allocation happens as soon as the method is called, and the rest of the line happens when execution reaches that point.



    The execution of the line actually has two stages: the first, the GetString method gets called. Second, its return value gets assigned to the local variable.



    As Marc Gravell pointed out, things can become more complex. The compiler may decide to simply not create the variable at all, or arrange things differently, as long as that produces the same result (something known as compiler optimizations). The rest of this answer assumes that the compiler didn't optimized anything and created a binary that matches exactly the given code, but in release builds we can expect some differences.



    *(and not captured by a lambda)






    share|improve this answer






















    • note - it is very possible (likely, even), that no stack space is actually reserved here; many local variables are elided completely by the compiler
      – Marc Gravell♦
      Aug 13 at 12:50










    • @MarcGravell Sure, in the case the compiler optimizes out the variable (probably in this case it does). For the simplicity sake, I'm assuming that the compiler compiles the code exactly as stands in the question, no optimizations involved. In real life, the actual binary might differ significantly from the code, as long as it produces the same result.
      – Alejandro
      Aug 13 at 13:06






    • 1




      One possible source of confusion: How can the compiler reserve memory at the start of the method when it doesn't even know how long the string is going to be? It might make sense to point out that space for the string content gets allocated somewhere within the GetString() call, and Method1 just stores a reference (to avoid the ugly word "pointer") to that space, with the amount of storage needed for the reference being always the same, independent of string length.
      – Guntram Blohm
      Aug 13 at 15:34















    up vote
    1
    down vote













    Memory allocation depends on how exactly the variable is used and where it's declared. In this case, being a local variable within a method*, memory is reserved as soon as the method is called (and not when execution reaches the place of its declaration), regardless of what could happen afterwards. So memory is set aside even if the client.GetString("http://msdn.microsoft.com") isn't called at all (couldn't happen at all here, but in more complex code that's possible).



    Note that you mention




    when the first line in Method1 is carried out




    This method has a single line, which consist in declaring a variable and assigning a value to it from calling another method. The fact that you've written it as two physical lines is irrelevant, as logically, your whole code consist of a single step. Again, variable "declaration" and allocation happens as soon as the method is called, and the rest of the line happens when execution reaches that point.



    The execution of the line actually has two stages: the first, the GetString method gets called. Second, its return value gets assigned to the local variable.



    As Marc Gravell pointed out, things can become more complex. The compiler may decide to simply not create the variable at all, or arrange things differently, as long as that produces the same result (something known as compiler optimizations). The rest of this answer assumes that the compiler didn't optimized anything and created a binary that matches exactly the given code, but in release builds we can expect some differences.



    *(and not captured by a lambda)






    share|improve this answer






















    • note - it is very possible (likely, even), that no stack space is actually reserved here; many local variables are elided completely by the compiler
      – Marc Gravell♦
      Aug 13 at 12:50










    • @MarcGravell Sure, in the case the compiler optimizes out the variable (probably in this case it does). For the simplicity sake, I'm assuming that the compiler compiles the code exactly as stands in the question, no optimizations involved. In real life, the actual binary might differ significantly from the code, as long as it produces the same result.
      – Alejandro
      Aug 13 at 13:06






    • 1




      One possible source of confusion: How can the compiler reserve memory at the start of the method when it doesn't even know how long the string is going to be? It might make sense to point out that space for the string content gets allocated somewhere within the GetString() call, and Method1 just stores a reference (to avoid the ugly word "pointer") to that space, with the amount of storage needed for the reference being always the same, independent of string length.
      – Guntram Blohm
      Aug 13 at 15:34













    up vote
    1
    down vote










    up vote
    1
    down vote









    Memory allocation depends on how exactly the variable is used and where it's declared. In this case, being a local variable within a method*, memory is reserved as soon as the method is called (and not when execution reaches the place of its declaration), regardless of what could happen afterwards. So memory is set aside even if the client.GetString("http://msdn.microsoft.com") isn't called at all (couldn't happen at all here, but in more complex code that's possible).



    Note that you mention




    when the first line in Method1 is carried out




    This method has a single line, which consist in declaring a variable and assigning a value to it from calling another method. The fact that you've written it as two physical lines is irrelevant, as logically, your whole code consist of a single step. Again, variable "declaration" and allocation happens as soon as the method is called, and the rest of the line happens when execution reaches that point.



    The execution of the line actually has two stages: the first, the GetString method gets called. Second, its return value gets assigned to the local variable.



    As Marc Gravell pointed out, things can become more complex. The compiler may decide to simply not create the variable at all, or arrange things differently, as long as that produces the same result (something known as compiler optimizations). The rest of this answer assumes that the compiler didn't optimized anything and created a binary that matches exactly the given code, but in release builds we can expect some differences.



    *(and not captured by a lambda)






    share|improve this answer














    Memory allocation depends on how exactly the variable is used and where it's declared. In this case, being a local variable within a method*, memory is reserved as soon as the method is called (and not when execution reaches the place of its declaration), regardless of what could happen afterwards. So memory is set aside even if the client.GetString("http://msdn.microsoft.com") isn't called at all (couldn't happen at all here, but in more complex code that's possible).



    Note that you mention




    when the first line in Method1 is carried out




    This method has a single line, which consist in declaring a variable and assigning a value to it from calling another method. The fact that you've written it as two physical lines is irrelevant, as logically, your whole code consist of a single step. Again, variable "declaration" and allocation happens as soon as the method is called, and the rest of the line happens when execution reaches that point.



    The execution of the line actually has two stages: the first, the GetString method gets called. Second, its return value gets assigned to the local variable.



    As Marc Gravell pointed out, things can become more complex. The compiler may decide to simply not create the variable at all, or arrange things differently, as long as that produces the same result (something known as compiler optimizations). The rest of this answer assumes that the compiler didn't optimized anything and created a binary that matches exactly the given code, but in release builds we can expect some differences.



    *(and not captured by a lambda)







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Aug 13 at 13:13

























    answered Aug 13 at 12:39









    Alejandro

    3,95711936




    3,95711936











    • note - it is very possible (likely, even), that no stack space is actually reserved here; many local variables are elided completely by the compiler
      – Marc Gravell♦
      Aug 13 at 12:50










    • @MarcGravell Sure, in the case the compiler optimizes out the variable (probably in this case it does). For the simplicity sake, I'm assuming that the compiler compiles the code exactly as stands in the question, no optimizations involved. In real life, the actual binary might differ significantly from the code, as long as it produces the same result.
      – Alejandro
      Aug 13 at 13:06






    • 1




      One possible source of confusion: How can the compiler reserve memory at the start of the method when it doesn't even know how long the string is going to be? It might make sense to point out that space for the string content gets allocated somewhere within the GetString() call, and Method1 just stores a reference (to avoid the ugly word "pointer") to that space, with the amount of storage needed for the reference being always the same, independent of string length.
      – Guntram Blohm
      Aug 13 at 15:34

















    • note - it is very possible (likely, even), that no stack space is actually reserved here; many local variables are elided completely by the compiler
      – Marc Gravell♦
      Aug 13 at 12:50










    • @MarcGravell Sure, in the case the compiler optimizes out the variable (probably in this case it does). For the simplicity sake, I'm assuming that the compiler compiles the code exactly as stands in the question, no optimizations involved. In real life, the actual binary might differ significantly from the code, as long as it produces the same result.
      – Alejandro
      Aug 13 at 13:06






    • 1




      One possible source of confusion: How can the compiler reserve memory at the start of the method when it doesn't even know how long the string is going to be? It might make sense to point out that space for the string content gets allocated somewhere within the GetString() call, and Method1 just stores a reference (to avoid the ugly word "pointer") to that space, with the amount of storage needed for the reference being always the same, independent of string length.
      – Guntram Blohm
      Aug 13 at 15:34
















    note - it is very possible (likely, even), that no stack space is actually reserved here; many local variables are elided completely by the compiler
    – Marc Gravell♦
    Aug 13 at 12:50




    note - it is very possible (likely, even), that no stack space is actually reserved here; many local variables are elided completely by the compiler
    – Marc Gravell♦
    Aug 13 at 12:50












    @MarcGravell Sure, in the case the compiler optimizes out the variable (probably in this case it does). For the simplicity sake, I'm assuming that the compiler compiles the code exactly as stands in the question, no optimizations involved. In real life, the actual binary might differ significantly from the code, as long as it produces the same result.
    – Alejandro
    Aug 13 at 13:06




    @MarcGravell Sure, in the case the compiler optimizes out the variable (probably in this case it does). For the simplicity sake, I'm assuming that the compiler compiles the code exactly as stands in the question, no optimizations involved. In real life, the actual binary might differ significantly from the code, as long as it produces the same result.
    – Alejandro
    Aug 13 at 13:06




    1




    1




    One possible source of confusion: How can the compiler reserve memory at the start of the method when it doesn't even know how long the string is going to be? It might make sense to point out that space for the string content gets allocated somewhere within the GetString() call, and Method1 just stores a reference (to avoid the ugly word "pointer") to that space, with the amount of storage needed for the reference being always the same, independent of string length.
    – Guntram Blohm
    Aug 13 at 15:34





    One possible source of confusion: How can the compiler reserve memory at the start of the method when it doesn't even know how long the string is going to be? It might make sense to point out that space for the string content gets allocated somewhere within the GetString() call, and Method1 just stores a reference (to avoid the ugly word "pointer") to that space, with the amount of storage needed for the reference being always the same, independent of string length.
    – Guntram Blohm
    Aug 13 at 15:34











    up vote
    0
    down vote













    Yes, in this case, the assignment happens immediately. Later on you'll learn about funny things like lazy loading, async and delegates and then you'll see instances where variables aren't initialised immediately. But don't worry, by the time you learn about those things, you'll know enough that it won't be difficult to understand.



    And congrats on learning the most beautiful programming language in the world.






    share|improve this answer
























      up vote
      0
      down vote













      Yes, in this case, the assignment happens immediately. Later on you'll learn about funny things like lazy loading, async and delegates and then you'll see instances where variables aren't initialised immediately. But don't worry, by the time you learn about those things, you'll know enough that it won't be difficult to understand.



      And congrats on learning the most beautiful programming language in the world.






      share|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Yes, in this case, the assignment happens immediately. Later on you'll learn about funny things like lazy loading, async and delegates and then you'll see instances where variables aren't initialised immediately. But don't worry, by the time you learn about those things, you'll know enough that it won't be difficult to understand.



        And congrats on learning the most beautiful programming language in the world.






        share|improve this answer












        Yes, in this case, the assignment happens immediately. Later on you'll learn about funny things like lazy loading, async and delegates and then you'll see instances where variables aren't initialised immediately. But don't worry, by the time you learn about those things, you'll know enough that it won't be difficult to understand.



        And congrats on learning the most beautiful programming language in the world.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Aug 13 at 12:46









        Captain Kenpachi

        4,12253049




        4,12253049












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