Does static make a difference for a const local variable?

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19















Imagine the following declaration:



void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



And a second one:



void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?










share|improve this question



















  • 3





    In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.

    – Matthieu Brucher
    Mar 1 at 10:31






  • 3





    out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)

    – user463035818
    Mar 1 at 10:35






  • 1





    @user463035818 I am having discussion during code review ;)

    – bartop
    Mar 1 at 10:37






  • 2





    depending on the reviewer that can be a real problem :P

    – user463035818
    Mar 1 at 10:38






  • 5





    @Scheff withoutStatic builds the array each times it is invoked from static data (.LC0). withStatic uses an array whose construction has been optimized as a constant (withStatic()::arr).

    – YSC
    Mar 1 at 10:53















19















Imagine the following declaration:



void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



And a second one:



void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?










share|improve this question



















  • 3





    In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.

    – Matthieu Brucher
    Mar 1 at 10:31






  • 3





    out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)

    – user463035818
    Mar 1 at 10:35






  • 1





    @user463035818 I am having discussion during code review ;)

    – bartop
    Mar 1 at 10:37






  • 2





    depending on the reviewer that can be a real problem :P

    – user463035818
    Mar 1 at 10:38






  • 5





    @Scheff withoutStatic builds the array each times it is invoked from static data (.LC0). withStatic uses an array whose construction has been optimized as a constant (withStatic()::arr).

    – YSC
    Mar 1 at 10:53













19












19








19


4






Imagine the following declaration:



void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



And a second one:



void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?










share|improve this question
















Imagine the following declaration:



void foo()
const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



And a second one:



void foo()
static const std::array<int, 80000> arr = /* a lot of different values*/;
//do stuff



What are the possible performance differences between these two if any? And is there any danger associated with any of these solutions?







c++ static const






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 1 at 15:57









Boann

37.4k1290121




37.4k1290121










asked Mar 1 at 10:27









bartopbartop

3,2601031




3,2601031







  • 3





    In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.

    – Matthieu Brucher
    Mar 1 at 10:31






  • 3





    out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)

    – user463035818
    Mar 1 at 10:35






  • 1





    @user463035818 I am having discussion during code review ;)

    – bartop
    Mar 1 at 10:37






  • 2





    depending on the reviewer that can be a real problem :P

    – user463035818
    Mar 1 at 10:38






  • 5





    @Scheff withoutStatic builds the array each times it is invoked from static data (.LC0). withStatic uses an array whose construction has been optimized as a constant (withStatic()::arr).

    – YSC
    Mar 1 at 10:53












  • 3





    In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.

    – Matthieu Brucher
    Mar 1 at 10:31






  • 3





    out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)

    – user463035818
    Mar 1 at 10:35






  • 1





    @user463035818 I am having discussion during code review ;)

    – bartop
    Mar 1 at 10:37






  • 2





    depending on the reviewer that can be a real problem :P

    – user463035818
    Mar 1 at 10:38






  • 5





    @Scheff withoutStatic builds the array each times it is invoked from static data (.LC0). withStatic uses an array whose construction has been optimized as a constant (withStatic()::arr).

    – YSC
    Mar 1 at 10:53







3




3





In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.

– Matthieu Brucher
Mar 1 at 10:31





In the static case they may not be on the stack, but in a read-only section. Probably compiler dependent as well.

– Matthieu Brucher
Mar 1 at 10:31




3




3





out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)

– user463035818
Mar 1 at 10:35





out of curiosity: do you have a real problem at hand, or is this just an academic exercise? (its a valid question in both cases)

– user463035818
Mar 1 at 10:35




1




1





@user463035818 I am having discussion during code review ;)

– bartop
Mar 1 at 10:37





@user463035818 I am having discussion during code review ;)

– bartop
Mar 1 at 10:37




2




2





depending on the reviewer that can be a real problem :P

– user463035818
Mar 1 at 10:38





depending on the reviewer that can be a real problem :P

– user463035818
Mar 1 at 10:38




5




5





@Scheff withoutStatic builds the array each times it is invoked from static data (.LC0). withStatic uses an array whose construction has been optimized as a constant (withStatic()::arr).

– YSC
Mar 1 at 10:53





@Scheff withoutStatic builds the array each times it is invoked from static data (.LC0). withStatic uses an array whose construction has been optimized as a constant (withStatic()::arr).

– YSC
Mar 1 at 10:53












4 Answers
4






active

oldest

votes


















12















And is there any danger associated with any of these solutions?




Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.



While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.






share|improve this answer
































    15














    Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.



    void f() 
    static const int x = get_x();
    // do something with x


    void g()
    const int x = get_x();
    // do something with x



    The difference between these two is that the first one will only call get_x() the first time that f() is called; x retains that value through the remainder of the program. The second one will call get_x() each time that g() is called.



    That matters if get_x() returns different values on subsequent calls:



    int current_x = 0;
    int get_x() return current_x++;





    share|improve this answer






























      5















      What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?




      The difference depends exactly on how you use foo().



      1st case:(low probability): Your implementation is such that you will call foo() only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.



      2nd case:(high probability): Your implementation is such that you will call foo() again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.






      share|improve this answer






























        2














        In this particular context, one point to consider regarding using static on a variable with initialization:



        From C++17 standard:




        6.7.1 Static storage duration [basic.stc.static]

        ...
        2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.







        share|improve this answer

























        • Actually, there are more differences: Its lifetime is different, for one....

          – CharonX
          Mar 1 at 14:19











        Your Answer






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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        12















        And is there any danger associated with any of these solutions?




        Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.



        While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.






        share|improve this answer





























          12















          And is there any danger associated with any of these solutions?




          Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.



          While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.






          share|improve this answer



























            12












            12








            12








            And is there any danger associated with any of these solutions?




            Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.



            While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.






            share|improve this answer
















            And is there any danger associated with any of these solutions?




            Non-static is dangerous because the array is huge, and the memory reserved for automatic storage is limited. Depending on the system and configuration, that array could use about 30% of the space available for automatic storage. As such, it greatly increases the possibility of stack overflow.



            While an optimiser might certainly avoid allocating memory on the stack, there are good reasons why you would want your non-optimised debug build to also not crash.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 1 at 11:00

























            answered Mar 1 at 10:53









            eerorikaeerorika

            87.9k663134




            87.9k663134























                15














                Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.



                void f() 
                static const int x = get_x();
                // do something with x


                void g()
                const int x = get_x();
                // do something with x



                The difference between these two is that the first one will only call get_x() the first time that f() is called; x retains that value through the remainder of the program. The second one will call get_x() each time that g() is called.



                That matters if get_x() returns different values on subsequent calls:



                int current_x = 0;
                int get_x() return current_x++;





                share|improve this answer



























                  15














                  Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.



                  void f() 
                  static const int x = get_x();
                  // do something with x


                  void g()
                  const int x = get_x();
                  // do something with x



                  The difference between these two is that the first one will only call get_x() the first time that f() is called; x retains that value through the remainder of the program. The second one will call get_x() each time that g() is called.



                  That matters if get_x() returns different values on subsequent calls:



                  int current_x = 0;
                  int get_x() return current_x++;





                  share|improve this answer

























                    15












                    15








                    15







                    Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.



                    void f() 
                    static const int x = get_x();
                    // do something with x


                    void g()
                    const int x = get_x();
                    // do something with x



                    The difference between these two is that the first one will only call get_x() the first time that f() is called; x retains that value through the remainder of the program. The second one will call get_x() each time that g() is called.



                    That matters if get_x() returns different values on subsequent calls:



                    int current_x = 0;
                    int get_x() return current_x++;





                    share|improve this answer













                    Forget the array for a moment. That muddles two separate issues. You've got answers that address the lifetime and storage issue. I'll address the initialization issue.



                    void f() 
                    static const int x = get_x();
                    // do something with x


                    void g()
                    const int x = get_x();
                    // do something with x



                    The difference between these two is that the first one will only call get_x() the first time that f() is called; x retains that value through the remainder of the program. The second one will call get_x() each time that g() is called.



                    That matters if get_x() returns different values on subsequent calls:



                    int current_x = 0;
                    int get_x() return current_x++;






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Mar 1 at 13:00









                    Pete BeckerPete Becker

                    58.7k442122




                    58.7k442122





















                        5















                        What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?




                        The difference depends exactly on how you use foo().



                        1st case:(low probability): Your implementation is such that you will call foo() only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.



                        2nd case:(high probability): Your implementation is such that you will call foo() again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.






                        share|improve this answer



























                          5















                          What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?




                          The difference depends exactly on how you use foo().



                          1st case:(low probability): Your implementation is such that you will call foo() only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.



                          2nd case:(high probability): Your implementation is such that you will call foo() again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.






                          share|improve this answer

























                            5












                            5








                            5








                            What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?




                            The difference depends exactly on how you use foo().



                            1st case:(low probability): Your implementation is such that you will call foo() only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.



                            2nd case:(high probability): Your implementation is such that you will call foo() again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.






                            share|improve this answer














                            What are the possible performance differences between these two if any?And is there any danger associated with any of these solutions?




                            The difference depends exactly on how you use foo().



                            1st case:(low probability): Your implementation is such that you will call foo() only once , maybe you have created separate function to divide code logic as practiced. Well in this case declaring as static is very bad, because a static variable or object remains in memory until programs ends . So just imagine that your variable occupying memory unnecessarily.



                            2nd case:(high probability): Your implementation is such that you will call foo() again and again . Then non-static object will get allocated and de allocated again and again.This will take huge amount of cpu clock cycles which is not desired .Use static in this case.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Mar 1 at 11:15









                            Abhishek GargAbhishek Garg

                            311110




                            311110





















                                2














                                In this particular context, one point to consider regarding using static on a variable with initialization:



                                From C++17 standard:




                                6.7.1 Static storage duration [basic.stc.static]

                                ...
                                2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.







                                share|improve this answer

























                                • Actually, there are more differences: Its lifetime is different, for one....

                                  – CharonX
                                  Mar 1 at 14:19















                                2














                                In this particular context, one point to consider regarding using static on a variable with initialization:



                                From C++17 standard:




                                6.7.1 Static storage duration [basic.stc.static]

                                ...
                                2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.







                                share|improve this answer

























                                • Actually, there are more differences: Its lifetime is different, for one....

                                  – CharonX
                                  Mar 1 at 14:19













                                2












                                2








                                2







                                In this particular context, one point to consider regarding using static on a variable with initialization:



                                From C++17 standard:




                                6.7.1 Static storage duration [basic.stc.static]

                                ...
                                2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.







                                share|improve this answer















                                In this particular context, one point to consider regarding using static on a variable with initialization:



                                From C++17 standard:




                                6.7.1 Static storage duration [basic.stc.static]

                                ...
                                2 If a variable with static storage duration has initialization or a destructor with side effects, it shall not be eliminated even if it appears to be unused, except that a class object or its copy/move may be eliminated as specified in 15.8.








                                share|improve this answer














                                share|improve this answer



                                share|improve this answer








                                edited Mar 1 at 14:23

























                                answered Mar 1 at 11:49









                                P.WP.W

                                17.5k41556




                                17.5k41556












                                • Actually, there are more differences: Its lifetime is different, for one....

                                  – CharonX
                                  Mar 1 at 14:19

















                                • Actually, there are more differences: Its lifetime is different, for one....

                                  – CharonX
                                  Mar 1 at 14:19
















                                Actually, there are more differences: Its lifetime is different, for one....

                                – CharonX
                                Mar 1 at 14:19





                                Actually, there are more differences: Its lifetime is different, for one....

                                – CharonX
                                Mar 1 at 14:19

















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