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std::is_member_function_pointer does not compile if false



2019 Community Moderator ElectionIs it possible to write a template to check for a function's existence?What does the explicit keyword mean?std::wstring VS std::stringWhy is “using namespace std” considered bad practice?How to print function pointers with cout?C++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming?Visual C++ 10.0 bug in std::reference_wrapper?std::async call of member functionstd::atomic_is_lock_free(shared_ptr<T>*) didn't compileCompiling an application for use in highly radioactive environmentsHow does “std::cout << std::endl;” compile?










9















What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:



template<class T> do_something() 
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value)
this->_t->x(); // _t is type of T*




What happens: The compiler does not compile if T does not bring the function. Small example:



#include <type_traits>
#include <iostream>

class Foo
public:
void x()
;

class Bar ;

int main()
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;



Compiler says:



is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;


So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.










share|improve this question






















  • you don't check if the type is a function pointer, you try to see a function pointer which did not exist

    – Klaus
    8 hours ago











  • Thanks for your comment! But how can I check that?

    – jagemue
    8 hours ago











  • stackoverflow.com/questions/257288/…

    – Klaus
    8 hours ago











  • Possible duplicate of Is it possible to write a template to check for a function's existence?

    – Klaus
    8 hours ago















9















What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:



template<class T> do_something() 
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value)
this->_t->x(); // _t is type of T*




What happens: The compiler does not compile if T does not bring the function. Small example:



#include <type_traits>
#include <iostream>

class Foo
public:
void x()
;

class Bar ;

int main()
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;



Compiler says:



is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;


So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.










share|improve this question






















  • you don't check if the type is a function pointer, you try to see a function pointer which did not exist

    – Klaus
    8 hours ago











  • Thanks for your comment! But how can I check that?

    – jagemue
    8 hours ago











  • stackoverflow.com/questions/257288/…

    – Klaus
    8 hours ago











  • Possible duplicate of Is it possible to write a template to check for a function's existence?

    – Klaus
    8 hours ago













9












9








9


2






What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:



template<class T> do_something() 
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value)
this->_t->x(); // _t is type of T*




What happens: The compiler does not compile if T does not bring the function. Small example:



#include <type_traits>
#include <iostream>

class Foo
public:
void x()
;

class Bar ;

int main()
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;



Compiler says:



is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;


So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.










share|improve this question














What I am looking for: I have a templated class and want to call a function if the class has the wanted function, something like:



template<class T> do_something() 
if constexpr (std::is_member_function_pointer<decltype(&T::x)>::value)
this->_t->x(); // _t is type of T*




What happens: The compiler does not compile if T does not bring the function. Small example:



#include <type_traits>
#include <iostream>

class Foo
public:
void x()
;

class Bar ;

int main()
std::cout << "Foo = " << std::is_member_function_pointer<decltype(&Foo::x)>::value << std::endl;
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;
return 0;



Compiler says:



is_member_function_pointer.cpp:17:69: error: no member named 'x' in 'Bar'; did you mean 'Foo::x'?
std::cout << "Bar = " << std::is_member_function_pointer<decltype(&Bar::x)>::value << std::endl;


So, what is the std::is_member_function_pointer for, when I can not use it in an if constexpr? If I just use this->_t->x() the compiler will fail, too, for sure.







c++ typetraits if-constexpr






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









jagemuejagemue

15919




15919












  • you don't check if the type is a function pointer, you try to see a function pointer which did not exist

    – Klaus
    8 hours ago











  • Thanks for your comment! But how can I check that?

    – jagemue
    8 hours ago











  • stackoverflow.com/questions/257288/…

    – Klaus
    8 hours ago











  • Possible duplicate of Is it possible to write a template to check for a function's existence?

    – Klaus
    8 hours ago

















  • you don't check if the type is a function pointer, you try to see a function pointer which did not exist

    – Klaus
    8 hours ago











  • Thanks for your comment! But how can I check that?

    – jagemue
    8 hours ago











  • stackoverflow.com/questions/257288/…

    – Klaus
    8 hours ago











  • Possible duplicate of Is it possible to write a template to check for a function's existence?

    – Klaus
    8 hours ago
















you don't check if the type is a function pointer, you try to see a function pointer which did not exist

– Klaus
8 hours ago





you don't check if the type is a function pointer, you try to see a function pointer which did not exist

– Klaus
8 hours ago













Thanks for your comment! But how can I check that?

– jagemue
8 hours ago





Thanks for your comment! But how can I check that?

– jagemue
8 hours ago













stackoverflow.com/questions/257288/…

– Klaus
8 hours ago





stackoverflow.com/questions/257288/…

– Klaus
8 hours ago













Possible duplicate of Is it possible to write a template to check for a function's existence?

– Klaus
8 hours ago





Possible duplicate of Is it possible to write a template to check for a function's existence?

– Klaus
8 hours ago












1 Answer
1






active

oldest

votes


















15














is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.



If you want to detect whether it exists or not, you can use the detection idiom. Example:



#include <experimental/type_traits>

template<class T>
using has_x = decltype(&T::x);

template<class T> void do_something(T t)
if constexpr (std::experimental::is_detected<has_x, T>::value)
t.x();



struct Foo
void x()
;

struct Bar ;

int main()
do_something(Foo);
do_something(Bar);



live example on godbolt.org




I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:



"checking expression validity in-place with C++17"






share|improve this answer






















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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15














    is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.



    If you want to detect whether it exists or not, you can use the detection idiom. Example:



    #include <experimental/type_traits>

    template<class T>
    using has_x = decltype(&T::x);

    template<class T> void do_something(T t)
    if constexpr (std::experimental::is_detected<has_x, T>::value)
    t.x();



    struct Foo
    void x()
    ;

    struct Bar ;

    int main()
    do_something(Foo);
    do_something(Bar);



    live example on godbolt.org




    I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:



    "checking expression validity in-place with C++17"






    share|improve this answer



























      15














      is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.



      If you want to detect whether it exists or not, you can use the detection idiom. Example:



      #include <experimental/type_traits>

      template<class T>
      using has_x = decltype(&T::x);

      template<class T> void do_something(T t)
      if constexpr (std::experimental::is_detected<has_x, T>::value)
      t.x();



      struct Foo
      void x()
      ;

      struct Bar ;

      int main()
      do_something(Foo);
      do_something(Bar);



      live example on godbolt.org




      I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:



      "checking expression validity in-place with C++17"






      share|improve this answer

























        15












        15








        15







        is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.



        If you want to detect whether it exists or not, you can use the detection idiom. Example:



        #include <experimental/type_traits>

        template<class T>
        using has_x = decltype(&T::x);

        template<class T> void do_something(T t)
        if constexpr (std::experimental::is_detected<has_x, T>::value)
        t.x();



        struct Foo
        void x()
        ;

        struct Bar ;

        int main()
        do_something(Foo);
        do_something(Bar);



        live example on godbolt.org




        I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:



        "checking expression validity in-place with C++17"






        share|improve this answer













        is_member_function_pointer doesn't detect the existence of an entity T::x, it assumes it does and returns whether or not it is a member function pointer.



        If you want to detect whether it exists or not, you can use the detection idiom. Example:



        #include <experimental/type_traits>

        template<class T>
        using has_x = decltype(&T::x);

        template<class T> void do_something(T t)
        if constexpr (std::experimental::is_detected<has_x, T>::value)
        t.x();



        struct Foo
        void x()
        ;

        struct Bar ;

        int main()
        do_something(Foo);
        do_something(Bar);



        live example on godbolt.org




        I have written an article on the general problem of checking the validity of an expression in different C++ Standard versions:



        "checking expression validity in-place with C++17"







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 8 hours ago









        Vittorio RomeoVittorio Romeo

        58.9k17161304




        58.9k17161304





























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