How to swap between functions implementation?












7














Is there any way to swap between two functions implementation in C++ ?



Something like this:



void printA(); // print a char
void printB(); // print b char

printA(); // output: a
printB(); // output: b

functionSwap(printA, printB);

printA(); // output: b
printB(); // output: a


I want to use it with the ExitProcess function.










share|improve this question




















  • 5




    This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping ExitProcess?
    – Andrey Akhmetov
    Nov 13 at 14:26










  • @AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
    – Mor Ben-Yosef
    Nov 13 at 14:31
















7














Is there any way to swap between two functions implementation in C++ ?



Something like this:



void printA(); // print a char
void printB(); // print b char

printA(); // output: a
printB(); // output: b

functionSwap(printA, printB);

printA(); // output: b
printB(); // output: a


I want to use it with the ExitProcess function.










share|improve this question




















  • 5




    This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping ExitProcess?
    – Andrey Akhmetov
    Nov 13 at 14:26










  • @AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
    – Mor Ben-Yosef
    Nov 13 at 14:31














7












7








7







Is there any way to swap between two functions implementation in C++ ?



Something like this:



void printA(); // print a char
void printB(); // print b char

printA(); // output: a
printB(); // output: b

functionSwap(printA, printB);

printA(); // output: b
printB(); // output: a


I want to use it with the ExitProcess function.










share|improve this question















Is there any way to swap between two functions implementation in C++ ?



Something like this:



void printA(); // print a char
void printB(); // print b char

printA(); // output: a
printB(); // output: b

functionSwap(printA, printB);

printA(); // output: b
printB(); // output: a


I want to use it with the ExitProcess function.







c++ c++11






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 at 14:29









Matthieu Brucher

11.8k22138




11.8k22138










asked Nov 13 at 14:24









Mor Ben-Yosef

517




517








  • 5




    This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping ExitProcess?
    – Andrey Akhmetov
    Nov 13 at 14:26










  • @AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
    – Mor Ben-Yosef
    Nov 13 at 14:31














  • 5




    This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping ExitProcess?
    – Andrey Akhmetov
    Nov 13 at 14:26










  • @AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
    – Mor Ben-Yosef
    Nov 13 at 14:31








5




5




This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping ExitProcess?
– Andrey Akhmetov
Nov 13 at 14:26




This seems to be a possible instance of the XY problem. What is the root goal you're trying to achieve by swapping ExitProcess?
– Andrey Akhmetov
Nov 13 at 14:26












@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31




@AndreyAkhmetov I am participating in some kind of competition in which everyone want to make the enemy process stop and there is the option of using DLL injection and using the ExitProcess function in order to make my process exit so I want to avoid the option of others to use the function in my process.
– Mor Ben-Yosef
Nov 13 at 14:31












2 Answers
2






active

oldest

votes


















16














You can bind a pointer to both functions in two variables and swap those.



void (*f1)() = printA;
void (*f2)() = printB;

f1(); // output: a
f2(); // output: b

std::swap(f1, f2);

f1(); // output: b
f2(); // output: a





share|improve this answer





















  • Don't use c function pointers, but std::function instead
    – hellow
    Nov 13 at 14:29






  • 13




    @hellow no. std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
    – Quentin
    Nov 13 at 14:32



















7














You need to wrap them in objects (or pointers to functions):



std::function<void()> myprintA = printA;
std::function<void()> myprintB = printB;

std::swap(myprintA, myprintB);

myprintA();
myprintB();


Otherwise, you are working with symbols themselves, and you can't swap this.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    16














    You can bind a pointer to both functions in two variables and swap those.



    void (*f1)() = printA;
    void (*f2)() = printB;

    f1(); // output: a
    f2(); // output: b

    std::swap(f1, f2);

    f1(); // output: b
    f2(); // output: a





    share|improve this answer





















    • Don't use c function pointers, but std::function instead
      – hellow
      Nov 13 at 14:29






    • 13




      @hellow no. std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
      – Quentin
      Nov 13 at 14:32
















    16














    You can bind a pointer to both functions in two variables and swap those.



    void (*f1)() = printA;
    void (*f2)() = printB;

    f1(); // output: a
    f2(); // output: b

    std::swap(f1, f2);

    f1(); // output: b
    f2(); // output: a





    share|improve this answer





















    • Don't use c function pointers, but std::function instead
      – hellow
      Nov 13 at 14:29






    • 13




      @hellow no. std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
      – Quentin
      Nov 13 at 14:32














    16












    16








    16






    You can bind a pointer to both functions in two variables and swap those.



    void (*f1)() = printA;
    void (*f2)() = printB;

    f1(); // output: a
    f2(); // output: b

    std::swap(f1, f2);

    f1(); // output: b
    f2(); // output: a





    share|improve this answer












    You can bind a pointer to both functions in two variables and swap those.



    void (*f1)() = printA;
    void (*f2)() = printB;

    f1(); // output: a
    f2(); // output: b

    std::swap(f1, f2);

    f1(); // output: b
    f2(); // output: a






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 13 at 14:28









    lubgr

    10.2k21745




    10.2k21745












    • Don't use c function pointers, but std::function instead
      – hellow
      Nov 13 at 14:29






    • 13




      @hellow no. std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
      – Quentin
      Nov 13 at 14:32


















    • Don't use c function pointers, but std::function instead
      – hellow
      Nov 13 at 14:29






    • 13




      @hellow no. std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
      – Quentin
      Nov 13 at 14:32
















    Don't use c function pointers, but std::function instead
    – hellow
    Nov 13 at 14:29




    Don't use c function pointers, but std::function instead
    – hellow
    Nov 13 at 14:29




    13




    13




    @hellow no. std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
    – Quentin
    Nov 13 at 14:32




    @hellow no. std::function is a heavywieght type-erased container for callable objects. If a function pointer is enough, use that.
    – Quentin
    Nov 13 at 14:32













    7














    You need to wrap them in objects (or pointers to functions):



    std::function<void()> myprintA = printA;
    std::function<void()> myprintB = printB;

    std::swap(myprintA, myprintB);

    myprintA();
    myprintB();


    Otherwise, you are working with symbols themselves, and you can't swap this.






    share|improve this answer


























      7














      You need to wrap them in objects (or pointers to functions):



      std::function<void()> myprintA = printA;
      std::function<void()> myprintB = printB;

      std::swap(myprintA, myprintB);

      myprintA();
      myprintB();


      Otherwise, you are working with symbols themselves, and you can't swap this.






      share|improve this answer
























        7












        7








        7






        You need to wrap them in objects (or pointers to functions):



        std::function<void()> myprintA = printA;
        std::function<void()> myprintB = printB;

        std::swap(myprintA, myprintB);

        myprintA();
        myprintB();


        Otherwise, you are working with symbols themselves, and you can't swap this.






        share|improve this answer












        You need to wrap them in objects (or pointers to functions):



        std::function<void()> myprintA = printA;
        std::function<void()> myprintB = printB;

        std::swap(myprintA, myprintB);

        myprintA();
        myprintB();


        Otherwise, you are working with symbols themselves, and you can't swap this.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 13 at 14:28









        Matthieu Brucher

        11.8k22138




        11.8k22138






























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