Allocation as default initialization
Clash Royale CLAN TAG#URR8PPP
Say I have a struct
(or class) with a dynamic array, its length, and a constructor:
struct array
int l;
int* t;
array(int length);
;
array::array(int length)
l=length;
t=new int[l];
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array
int l;
int* t = new int[l];
array(int length);
array::array(int length)
l=length;
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
add a comment |
Say I have a struct
(or class) with a dynamic array, its length, and a constructor:
struct array
int l;
int* t;
array(int length);
;
array::array(int length)
l=length;
t=new int[l];
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array
int l;
int* t = new int[l];
array(int length);
array::array(int length)
l=length;
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuseI
,l
and1
. :-)
– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
add a comment |
Say I have a struct
(or class) with a dynamic array, its length, and a constructor:
struct array
int l;
int* t;
array(int length);
;
array::array(int length)
l=length;
t=new int[l];
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array
int l;
int* t = new int[l];
array(int length);
array::array(int length)
l=length;
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
Say I have a struct
(or class) with a dynamic array, its length, and a constructor:
struct array
int l;
int* t;
array(int length);
;
array::array(int length)
l=length;
t=new int[l];
I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:
struct array
int l;
int* t = new int[l];
array(int length);
array::array(int length)
l=length;
It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.
c++ constructor
c++ constructor
asked Jan 28 at 10:24
tarulentarulen
1,798413
1,798413
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuseI
,l
and1
. :-)
– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
add a comment |
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuseI
,l
and1
. :-)
– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
7
7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse
I
, l
and 1
. :-)– Heinzi
Jan 28 at 14:37
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse
I
, l
and 1
. :-)– Heinzi
Jan 28 at 14:37
5
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09
add a comment |
2 Answers
2
active
oldest
votes
This code is not correct.
int* t = new int[l];
will happen before l=length;
, thus reading the uninitialized variable l
. Member initializers are handled before the constructor's body runs.
array::array(int length) : llength
instead would work because l
is declared before t
.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector
.
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array
, when t
is initialized l
is not initialized yet. For objects with automatic and dynamic storage duration l
will be initialized to indeterminate value, then the usage of l
(i.e. new int[l]
) leads to UB.
Note that l=length;
inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l])
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectsl
will get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This code is not correct.
int* t = new int[l];
will happen before l=length;
, thus reading the uninitialized variable l
. Member initializers are handled before the constructor's body runs.
array::array(int length) : llength
instead would work because l
is declared before t
.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector
.
add a comment |
This code is not correct.
int* t = new int[l];
will happen before l=length;
, thus reading the uninitialized variable l
. Member initializers are handled before the constructor's body runs.
array::array(int length) : llength
instead would work because l
is declared before t
.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector
.
add a comment |
This code is not correct.
int* t = new int[l];
will happen before l=length;
, thus reading the uninitialized variable l
. Member initializers are handled before the constructor's body runs.
array::array(int length) : llength
instead would work because l
is declared before t
.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector
.
This code is not correct.
int* t = new int[l];
will happen before l=length;
, thus reading the uninitialized variable l
. Member initializers are handled before the constructor's body runs.
array::array(int length) : llength
instead would work because l
is declared before t
.
However, doing this "by hand" is a bad idea to begin with. You should be using std::vector
.
answered Jan 28 at 10:29
Baum mit AugenBaum mit Augen
41.2k12118154
41.2k12118154
add a comment |
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array
, when t
is initialized l
is not initialized yet. For objects with automatic and dynamic storage duration l
will be initialized to indeterminate value, then the usage of l
(i.e. new int[l]
) leads to UB.
Note that l=length;
inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l])
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectsl
will get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array
, when t
is initialized l
is not initialized yet. For objects with automatic and dynamic storage duration l
will be initialized to indeterminate value, then the usage of l
(i.e. new int[l]
) leads to UB.
Note that l=length;
inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l])
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectsl
will get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
add a comment |
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array
, when t
is initialized l
is not initialized yet. For objects with automatic and dynamic storage duration l
will be initialized to indeterminate value, then the usage of l
(i.e. new int[l]
) leads to UB.
Note that l=length;
inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l])
The 2nd code snippet might have undefined behavior.
The data members are initialized at the order of how they're declared. For class array
, when t
is initialized l
is not initialized yet. For objects with automatic and dynamic storage duration l
will be initialized to indeterminate value, then the usage of l
(i.e. new int[l]
) leads to UB.
Note that l=length;
inside the body of the constructor is just assignment; the initialization of data members has been finished before that.
BTW: With member initializer list the 1st code snippet chould be rewritten as
array::array(int length) : l(length), t(new int[l])
edited Jan 28 at 10:33
answered Jan 28 at 10:28
songyuanyaosongyuanyao
91.6k11175238
91.6k11175238
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectsl
will get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
add a comment |
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
@Ruslan Because for static or thread-local objectsl
will get zero initialized at first.
– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
3
3
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
Why do you say "might"? It does have UB, unconditionally.
– Ruslan
Jan 28 at 16:29
1
1
@Ruslan Because for static or thread-local objects
l
will get zero initialized at first.– songyuanyao
Jan 29 at 1:13
@Ruslan Because for static or thread-local objects
l
will get zero initialized at first.– songyuanyao
Jan 29 at 1:13
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
@Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)
– Lightness Races in Orbit
Jan 30 at 16:33
add a comment |
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7
don't use 'l' as a variable name, It's easily confused with the number '1',
– regomodo
Jan 28 at 13:31
@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse
I
,l
and1
. :-)– Heinzi
Jan 28 at 14:37
5
id rather people read "Clean Code" instead
– regomodo
Jan 28 at 14:54
2
@Heinzi Or maybe stop using single letter variables damnit.
– Bakuriu
Jan 28 at 20:09