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Removing class pointer from member function pointer type

C++ function typesHow to call a parent class function from derived class function?error: request for member '..' in '..' which is of non-class typePretty-print C++ STL containersC++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming?What's the difference between std::move and std::forwardWhy is my program slow when looping over exactly 8192 elements?Partial template class specialization with member function pointerReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsInstantiating a templated class from a templated member function of a not templated classInconsistent parameter pack deduction int and int& in variadic templated member function that creates a thread that runs a member function

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6

2

I saw this code snippet in the example on cppreference for std::is_function and I don't understand how it works. Could someone explain to me why U gets deduced as it does in PM_traits?

struct A 
 int fun() const&;
;

template<typename>
struct PM_traits ; 

template<class T, class U>
struct PM_traits<U T::*> 
 using member_type = U;
;

int main() 
 using T = PM_traits<decltype(&A::fun)>::member_type; // T is int() const&

c++ c++11 templates

share|improve this question

asked 8 hours ago

willtunnelswilltunnels

3122 bronze badges

New contributor

willtunnels is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I've verified that this is the deduced type with g++. T is in fact int() const&.
  
  – willtunnels
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, it seems you are right. But I have a hard time understanding what a int() const& can possibly mean outside the context of a member function. const and the ref qualifier & are meaningless (and forbidden) on a free function.
  
  – François Andrieux
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @FrançoisAndrieux you can simply it, by making a simple member function, non-const. It would still be a function type.
  
  – SergeyA
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @FrançoisAndrieux Yes, they are forbidden on free functions, but not on function types in general. Remember that it's legal to declare (not define) a member function using a typedef, for example. Live example
  
  – Angew
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Angew That's fascinating, thank you for the example.
  
  – François Andrieux
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

6

2

I saw this code snippet in the example on cppreference for std::is_function and I don't understand how it works. Could someone explain to me why U gets deduced as it does in PM_traits?

struct A 
 int fun() const&;
;

template<typename>
struct PM_traits ; 

template<class T, class U>
struct PM_traits<U T::*> 
 using member_type = U;
;

int main() 
 using T = PM_traits<decltype(&A::fun)>::member_type; // T is int() const&

c++ c++11 templates

share|improve this question

asked 8 hours ago

willtunnelswilltunnels

3122 bronze badges

New contributor

willtunnels is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I've verified that this is the deduced type with g++. T is in fact int() const&.
  
  – willtunnels
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Well, it seems you are right. But I have a hard time understanding what a int() const& can possibly mean outside the context of a member function. const and the ref qualifier & are meaningless (and forbidden) on a free function.
  
  – François Andrieux
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @FrançoisAndrieux you can simply it, by making a simple member function, non-const. It would still be a function type.
  
  – SergeyA
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @FrançoisAndrieux Yes, they are forbidden on free functions, but not on function types in general. Remember that it's legal to declare (not define) a member function using a typedef, for example. Live example
  
  – Angew
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Angew That's fascinating, thank you for the example.
  
  – François Andrieux
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

I saw this code snippet in the example on cppreference for std::is_function and I don't understand how it works. Could someone explain to me why U gets deduced as it does in PM_traits?

struct A 
 int fun() const&;
;

template<typename>
struct PM_traits ; 

template<class T, class U>
struct PM_traits<U T::*> 
 using member_type = U;
;

int main() 
 using T = PM_traits<decltype(&A::fun)>::member_type; // T is int() const&

c++ c++11 templates

share|improve this question

asked 8 hours ago

willtunnelswilltunnels

3122 bronze badges

New contributor

willtunnels is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

struct A 
 int fun() const&;
;

template<typename>
struct PM_traits ; 

template<class T, class U>
struct PM_traits<U T::*> 
 using member_type = U;
;

int main() 
 using T = PM_traits<decltype(&A::fun)>::member_type; // T is int() const&

share|improve this question

asked 8 hours ago

willtunnelswilltunnels

3122 bronze badges

New contributor

willtunnels is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 8 hours ago

willtunnelswilltunnels

3122 bronze badges

New contributor

willtunnels is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

willtunnelswilltunnels

3122 bronze badges

New contributor

willtunnels is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

willtunnelswilltunnels

3122 bronze badges

1 Answer
1

active

oldest

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9

U T::* is a type such that when we have U T::* p, p points to a member of class T, and that member is of type U.

fun is a function of type int () const &: a const &-qualified function taking no parameters and returning int, and it's a member of class A. Therefore, in the deduction, T is deduced to A and U is deduced to the type of A::fun, which is int () const &.

---

It may look a bit confusing, because if the type of &A::fun was spelled out explicitly, it would have to be written int (A::*)() const &. However, in the template's case, the type int () const & is "hidden" behind the name U, so the pointer to member is then just U A::*. It's similar to how type names can be used to simplify the syntax of normal function pointers:

int foo(char, double) return 42; 

using Fun = int (char, double);
Fun *p = &foo;
// instead of:
int (*q)(char, double) = &foo;

The same happens in the template, just with A::* instead of *.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

AngewAngew

138k1111 gold badges273273 silver badges363363 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I can write int(A::* p)() const & = &A::fun; but int () const & A::* p = &A::fun; gives me an error. Why is there a discrepancy between what happens here and what happens in templates?
  
  – willtunnels
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @willtunnels Because that's how C++ syntax works. Notice that you can write using V = int () const &; V A::* p = &A::fun;. It's the same as writing const int *p; vs. using T = int*; const T p;.
  
  – Angew
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That generally seems to work except when ref qualifiers are introduced. I put in your code exactly and g++ spat out cannot convert 'int (A::*)() const &' to 'int (A::*)() const' in initialization. Am I missing something obvious? Can someone confirm that that code actually works for them? (Thanks for your help by the way.)
  
  – willtunnels
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So, based on some experimentation, it looks like clang handles this correctly (?) but gcc fails.
  
  – willtunnels
  5 hours ago
  

  
  
  
  

add a comment | 

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9

U T::* is a type such that when we have U T::* p, p points to a member of class T, and that member is of type U.

fun is a function of type int () const &: a const &-qualified function taking no parameters and returning int, and it's a member of class A. Therefore, in the deduction, T is deduced to A and U is deduced to the type of A::fun, which is int () const &.

---

It may look a bit confusing, because if the type of &A::fun was spelled out explicitly, it would have to be written int (A::*)() const &. However, in the template's case, the type int () const & is "hidden" behind the name U, so the pointer to member is then just U A::*. It's similar to how type names can be used to simplify the syntax of normal function pointers:

int foo(char, double) return 42; 

using Fun = int (char, double);
Fun *p = &foo;
// instead of:
int (*q)(char, double) = &foo;

The same happens in the template, just with A::* instead of *.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

AngewAngew

138k1111 gold badges273273 silver badges363363 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I can write int(A::* p)() const & = &A::fun; but int () const & A::* p = &A::fun; gives me an error. Why is there a discrepancy between what happens here and what happens in templates?
  
  – willtunnels
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @willtunnels Because that's how C++ syntax works. Notice that you can write using V = int () const &; V A::* p = &A::fun;. It's the same as writing const int *p; vs. using T = int*; const T p;.
  
  – Angew
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That generally seems to work except when ref qualifiers are introduced. I put in your code exactly and g++ spat out cannot convert 'int (A::*)() const &' to 'int (A::*)() const' in initialization. Am I missing something obvious? Can someone confirm that that code actually works for them? (Thanks for your help by the way.)
  
  – willtunnels
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So, based on some experimentation, it looks like clang handles this correctly (?) but gcc fails.
  
  – willtunnels
  5 hours ago
  

  
  
  
  

add a comment | 

9

U T::* is a type such that when we have U T::* p, p points to a member of class T, and that member is of type U.

fun is a function of type int () const &: a const &-qualified function taking no parameters and returning int, and it's a member of class A. Therefore, in the deduction, T is deduced to A and U is deduced to the type of A::fun, which is int () const &.

---

It may look a bit confusing, because if the type of &A::fun was spelled out explicitly, it would have to be written int (A::*)() const &. However, in the template's case, the type int () const & is "hidden" behind the name U, so the pointer to member is then just U A::*. It's similar to how type names can be used to simplify the syntax of normal function pointers:

int foo(char, double) return 42; 

using Fun = int (char, double);
Fun *p = &foo;
// instead of:
int (*q)(char, double) = &foo;

The same happens in the template, just with A::* instead of *.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

AngewAngew

138k1111 gold badges273273 silver badges363363 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I can write int(A::* p)() const & = &A::fun; but int () const & A::* p = &A::fun; gives me an error. Why is there a discrepancy between what happens here and what happens in templates?
  
  – willtunnels
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @willtunnels Because that's how C++ syntax works. Notice that you can write using V = int () const &; V A::* p = &A::fun;. It's the same as writing const int *p; vs. using T = int*; const T p;.
  
  – Angew
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That generally seems to work except when ref qualifiers are introduced. I put in your code exactly and g++ spat out cannot convert 'int (A::*)() const &' to 'int (A::*)() const' in initialization. Am I missing something obvious? Can someone confirm that that code actually works for them? (Thanks for your help by the way.)
  
  – willtunnels
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So, based on some experimentation, it looks like clang handles this correctly (?) but gcc fails.
  
  – willtunnels
  5 hours ago
  

  
  
  
  

add a comment | 

U T::* is a type such that when we have U T::* p, p points to a member of class T, and that member is of type U.

fun is a function of type int () const &: a const &-qualified function taking no parameters and returning int, and it's a member of class A. Therefore, in the deduction, T is deduced to A and U is deduced to the type of A::fun, which is int () const &.

---

It may look a bit confusing, because if the type of &A::fun was spelled out explicitly, it would have to be written int (A::*)() const &. However, in the template's case, the type int () const & is "hidden" behind the name U, so the pointer to member is then just U A::*. It's similar to how type names can be used to simplify the syntax of normal function pointers:

int foo(char, double) return 42; 

using Fun = int (char, double);
Fun *p = &foo;
// instead of:
int (*q)(char, double) = &foo;

The same happens in the template, just with A::* instead of *.

share|improve this answer

edited 7 hours ago

answered 8 hours ago

AngewAngew

138k1111 gold badges273273 silver badges363363 bronze badges

int foo(char, double) return 42; 

using Fun = int (char, double);
Fun *p = &foo;
// instead of:
int (*q)(char, double) = &foo;

share|improve this answer

edited 7 hours ago

answered 8 hours ago

AngewAngew

138k1111 gold badges273273 silver badges363363 bronze badges

answered 8 hours ago

AngewAngew

138k1111 gold badges273273 silver badges363363 bronze badges

answered 8 hours ago

AngewAngew

138k1111 gold badges273273 silver badges363363 bronze badges