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std::is_member_function_pointer does not compile if false

Is 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

1

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

asked Mar 14 at 11:52

jagemuejagemue

159110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you don't check if the type is a function pointer, you try to see a function pointer which did not exist
  
  – Klaus
  Mar 14 at 11:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for your comment! But how can I check that?
  
  – jagemue
  Mar 14 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stackoverflow.com/questions/257288/…
  
  – Klaus
  Mar 14 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Is it possible to write a template to check for a function's existence?
  
  – Klaus
  Mar 14 at 11:56
  

  
  
  
  

add a comment | 

9

1

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

asked Mar 14 at 11:52

jagemuejagemue

159110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you don't check if the type is a function pointer, you try to see a function pointer which did not exist
  
  – Klaus
  Mar 14 at 11:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for your comment! But how can I check that?
  
  – jagemue
  Mar 14 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stackoverflow.com/questions/257288/…
  
  – Klaus
  Mar 14 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Is it possible to write a template to check for a function's existence?
  
  – Klaus
  Mar 14 at 11:56
  

  
  
  
  

add a comment | 

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

asked Mar 14 at 11:52

jagemuejagemue

159110

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

#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;

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;

share|improve this question

asked Mar 14 at 11:52

jagemuejagemue

159110

share|improve this question

asked Mar 14 at 11:52

jagemuejagemue

159110

asked Mar 14 at 11:52

jagemuejagemue

159110

asked Mar 14 at 11:52

jagemuejagemue

159110

1 Answer
1

active

oldest

votes

16

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

answered Mar 14 at 11:56

Vittorio RomeoVittorio Romeo

59k17161305

add a comment | 

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16

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

answered Mar 14 at 11:56

Vittorio RomeoVittorio Romeo

59k17161305

add a comment | 

16

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

answered Mar 14 at 11:56

Vittorio RomeoVittorio Romeo

59k17161305

add a comment | 

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

answered Mar 14 at 11:56

Vittorio RomeoVittorio Romeo

59k17161305

#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);

share|improve this answer

answered Mar 14 at 11:56

Vittorio RomeoVittorio Romeo

59k17161305

answered Mar 14 at 11:56

Vittorio RomeoVittorio Romeo

59k17161305

answered Mar 14 at 11:56

Vittorio RomeoVittorio Romeo

59k17161305