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How to call a function with default parameter through a pointer to function that is the return of another function?



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6















I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.




  • Mult's second parameter is default So when I call Double, Double returns the address of Mult thus I can pass only one argument.

But It doesn't work with pointer to function:



int Mult(int x, int y = 2) // y is default
return x * y;


using pFn = int(*)(int, int);


pFn Double(int x)
return Mult;


int main(int argc, char* argv[])

pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default




Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.










share|improve this question



















  • 1





    What is the point of Double taking an integer parameter that it doesn't use?

    – scohe001
    32 mins ago






  • 1





    Similar: Howto: c++ Function Pointer with default values

    – TrebledJ
    32 mins ago


















6















I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.




  • Mult's second parameter is default So when I call Double, Double returns the address of Mult thus I can pass only one argument.

But It doesn't work with pointer to function:



int Mult(int x, int y = 2) // y is default
return x * y;


using pFn = int(*)(int, int);


pFn Double(int x)
return Mult;


int main(int argc, char* argv[])

pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default




Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.










share|improve this question



















  • 1





    What is the point of Double taking an integer parameter that it doesn't use?

    – scohe001
    32 mins ago






  • 1





    Similar: Howto: c++ Function Pointer with default values

    – TrebledJ
    32 mins ago














6












6








6


3






I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.




  • Mult's second parameter is default So when I call Double, Double returns the address of Mult thus I can pass only one argument.

But It doesn't work with pointer to function:



int Mult(int x, int y = 2) // y is default
return x * y;


using pFn = int(*)(int, int);


pFn Double(int x)
return Mult;


int main(int argc, char* argv[])

pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default




Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.










share|improve this question
















I have a function Mult that takes two integers and returns the product of its parameters. And a function Double that takes an integer and returns a pointer to function that returns an integer and takes two integer parameters like Mult.




  • Mult's second parameter is default So when I call Double, Double returns the address of Mult thus I can pass only one argument.

But It doesn't work with pointer to function:



int Mult(int x, int y = 2) // y is default
return x * y;


using pFn = int(*)(int, int);


pFn Double(int x)
return Mult;


int main(int argc, char* argv[])

pFn func = Double(0);
cout << func(7, 4) << endl; // ok
//cout << func(7) << endl; // error: Too few arguments
cout << Mult(4) << endl; // ok. the second argument is default




Above if I call Mult with a single argument it works fine because the second argument is default but calling it through the pointer func it fails. func is pointer to function that takes two integers and returns an int.







c++ function-pointers default-arguments






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 28 mins ago









ShadowRanger

64.1k661100




64.1k661100










asked 35 mins ago









Syfu_HSyfu_H

1506




1506







  • 1





    What is the point of Double taking an integer parameter that it doesn't use?

    – scohe001
    32 mins ago






  • 1





    Similar: Howto: c++ Function Pointer with default values

    – TrebledJ
    32 mins ago













  • 1





    What is the point of Double taking an integer parameter that it doesn't use?

    – scohe001
    32 mins ago






  • 1





    Similar: Howto: c++ Function Pointer with default values

    – TrebledJ
    32 mins ago








1




1





What is the point of Double taking an integer parameter that it doesn't use?

– scohe001
32 mins ago





What is the point of Double taking an integer parameter that it doesn't use?

– scohe001
32 mins ago




1




1





Similar: Howto: c++ Function Pointer with default values

– TrebledJ
32 mins ago






Similar: Howto: c++ Function Pointer with default values

– TrebledJ
32 mins ago













2 Answers
2






active

oldest

votes


















9














Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).



The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.






share|improve this answer






























    1














    For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:



    auto Double() 
    return [](int x,int y=2) return Mult(x,y); ;



    And by using a variadic lambda (thanks to @Artyer) you do not even have to repeat the default value:



    #include <iostream>

    int Mult(int x, int y = 2) // y is default
    return x * y;


    auto Double()
    return [](auto... args) return Mult(args...); ;


    int main(int argc, char* argv[])
    auto func = Double();
    std::cout << func(7, 4) << 'n'; // ok
    std::cout << func(7) << 'n'; // ok
    std::cout << Mult(4) << 'n'; // ok



    Live demo






    share|improve this answer

























    • Note that this involves repeating the default explicitly inside Double when defining the lambda, which limits the utility significantly.

      – ShadowRanger
      15 mins ago











    • @ShadowRanger yes, added a note

      – user463035818
      10 mins ago











    • To not have to repeat the defaults, just forward variadic arguments: return [](auto... args) return Mult(args...); . Or with perfect forwarding (Which is not really necessary here because this just copies ints, but may be for other functions) return [](auto&&... args) noexcept(noexcept(Mult(std::forward<decltype(args)>(args)...))) -> decltype(auto) return Mult(std::forward<decltype(args)>(args)...); ;

      – Artyer
      6 mins ago











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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9














    Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).



    The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.






    share|improve this answer



























      9














      Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).



      The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.






      share|improve this answer

























        9












        9








        9







        Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).



        The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.






        share|improve this answer













        Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).



        The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 31 mins ago









        ShadowRangerShadowRanger

        64.1k661100




        64.1k661100























            1














            For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:



            auto Double() 
            return [](int x,int y=2) return Mult(x,y); ;



            And by using a variadic lambda (thanks to @Artyer) you do not even have to repeat the default value:



            #include <iostream>

            int Mult(int x, int y = 2) // y is default
            return x * y;


            auto Double()
            return [](auto... args) return Mult(args...); ;


            int main(int argc, char* argv[])
            auto func = Double();
            std::cout << func(7, 4) << 'n'; // ok
            std::cout << func(7) << 'n'; // ok
            std::cout << Mult(4) << 'n'; // ok



            Live demo






            share|improve this answer

























            • Note that this involves repeating the default explicitly inside Double when defining the lambda, which limits the utility significantly.

              – ShadowRanger
              15 mins ago











            • @ShadowRanger yes, added a note

              – user463035818
              10 mins ago











            • To not have to repeat the defaults, just forward variadic arguments: return [](auto... args) return Mult(args...); . Or with perfect forwarding (Which is not really necessary here because this just copies ints, but may be for other functions) return [](auto&&... args) noexcept(noexcept(Mult(std::forward<decltype(args)>(args)...))) -> decltype(auto) return Mult(std::forward<decltype(args)>(args)...); ;

              – Artyer
              6 mins ago















            1














            For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:



            auto Double() 
            return [](int x,int y=2) return Mult(x,y); ;



            And by using a variadic lambda (thanks to @Artyer) you do not even have to repeat the default value:



            #include <iostream>

            int Mult(int x, int y = 2) // y is default
            return x * y;


            auto Double()
            return [](auto... args) return Mult(args...); ;


            int main(int argc, char* argv[])
            auto func = Double();
            std::cout << func(7, 4) << 'n'; // ok
            std::cout << func(7) << 'n'; // ok
            std::cout << Mult(4) << 'n'; // ok



            Live demo






            share|improve this answer

























            • Note that this involves repeating the default explicitly inside Double when defining the lambda, which limits the utility significantly.

              – ShadowRanger
              15 mins ago











            • @ShadowRanger yes, added a note

              – user463035818
              10 mins ago











            • To not have to repeat the defaults, just forward variadic arguments: return [](auto... args) return Mult(args...); . Or with perfect forwarding (Which is not really necessary here because this just copies ints, but may be for other functions) return [](auto&&... args) noexcept(noexcept(Mult(std::forward<decltype(args)>(args)...))) -> decltype(auto) return Mult(std::forward<decltype(args)>(args)...); ;

              – Artyer
              6 mins ago













            1












            1








            1







            For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:



            auto Double() 
            return [](int x,int y=2) return Mult(x,y); ;



            And by using a variadic lambda (thanks to @Artyer) you do not even have to repeat the default value:



            #include <iostream>

            int Mult(int x, int y = 2) // y is default
            return x * y;


            auto Double()
            return [](auto... args) return Mult(args...); ;


            int main(int argc, char* argv[])
            auto func = Double();
            std::cout << func(7, 4) << 'n'; // ok
            std::cout << func(7) << 'n'; // ok
            std::cout << Mult(4) << 'n'; // ok



            Live demo






            share|improve this answer















            For the "why not" I refer you to this answer. If you want to somehow keep the ability to use a default, you need to provide something more than a function pointer, eg a lamdba will do:



            auto Double() 
            return [](int x,int y=2) return Mult(x,y); ;



            And by using a variadic lambda (thanks to @Artyer) you do not even have to repeat the default value:



            #include <iostream>

            int Mult(int x, int y = 2) // y is default
            return x * y;


            auto Double()
            return [](auto... args) return Mult(args...); ;


            int main(int argc, char* argv[])
            auto func = Double();
            std::cout << func(7, 4) << 'n'; // ok
            std::cout << func(7) << 'n'; // ok
            std::cout << Mult(4) << 'n'; // ok



            Live demo







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 4 mins ago

























            answered 18 mins ago









            user463035818user463035818

            19.3k42971




            19.3k42971












            • Note that this involves repeating the default explicitly inside Double when defining the lambda, which limits the utility significantly.

              – ShadowRanger
              15 mins ago











            • @ShadowRanger yes, added a note

              – user463035818
              10 mins ago











            • To not have to repeat the defaults, just forward variadic arguments: return [](auto... args) return Mult(args...); . Or with perfect forwarding (Which is not really necessary here because this just copies ints, but may be for other functions) return [](auto&&... args) noexcept(noexcept(Mult(std::forward<decltype(args)>(args)...))) -> decltype(auto) return Mult(std::forward<decltype(args)>(args)...); ;

              – Artyer
              6 mins ago

















            • Note that this involves repeating the default explicitly inside Double when defining the lambda, which limits the utility significantly.

              – ShadowRanger
              15 mins ago











            • @ShadowRanger yes, added a note

              – user463035818
              10 mins ago











            • To not have to repeat the defaults, just forward variadic arguments: return [](auto... args) return Mult(args...); . Or with perfect forwarding (Which is not really necessary here because this just copies ints, but may be for other functions) return [](auto&&... args) noexcept(noexcept(Mult(std::forward<decltype(args)>(args)...))) -> decltype(auto) return Mult(std::forward<decltype(args)>(args)...); ;

              – Artyer
              6 mins ago
















            Note that this involves repeating the default explicitly inside Double when defining the lambda, which limits the utility significantly.

            – ShadowRanger
            15 mins ago





            Note that this involves repeating the default explicitly inside Double when defining the lambda, which limits the utility significantly.

            – ShadowRanger
            15 mins ago













            @ShadowRanger yes, added a note

            – user463035818
            10 mins ago





            @ShadowRanger yes, added a note

            – user463035818
            10 mins ago













            To not have to repeat the defaults, just forward variadic arguments: return [](auto... args) return Mult(args...); . Or with perfect forwarding (Which is not really necessary here because this just copies ints, but may be for other functions) return [](auto&&... args) noexcept(noexcept(Mult(std::forward<decltype(args)>(args)...))) -> decltype(auto) return Mult(std::forward<decltype(args)>(args)...); ;

            – Artyer
            6 mins ago





            To not have to repeat the defaults, just forward variadic arguments: return [](auto... args) return Mult(args...); . Or with perfect forwarding (Which is not really necessary here because this just copies ints, but may be for other functions) return [](auto&&... args) noexcept(noexcept(Mult(std::forward<decltype(args)>(args)...))) -> decltype(auto) return Mult(std::forward<decltype(args)>(args)...); ;

            – Artyer
            6 mins ago

















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