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7

demo:

#include<iostream>
struct A int i = 10; ;
struct B : A ;

int main()
 std::cout << "decltype(&B::i) == int A::* ? " << std::boolalpha
 << std::is_same<decltype(&B::i), int A::*>::value << 'n'; //#1
 A a;
 std::cout << a.*(&A::i) << 'n';

 std::cout << "decltype(&B::i) == int B::* ? "
 << std::is_same<decltype(&B::i), int B::*>::value << 'n'; //#2
 B b;
 std::cout << b.*(&B::i) << 'n';

The code prints

decltype(&B::i) == int A::* ? true
10
decltype(&B::i) == int B::* ? false
10

I used the example in [expr.unary.op]/3, where the standard says that the type of &B::i is int A::*, but that is not normative.

c++ language-lawyer pointer-to-member

share|improve this question

edited 14 mins ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

add a comment | 

7

demo:

#include<iostream>
struct A int i = 10; ;
struct B : A ;

int main()
 std::cout << "decltype(&B::i) == int A::* ? " << std::boolalpha
 << std::is_same<decltype(&B::i), int A::*>::value << 'n'; //#1
 A a;
 std::cout << a.*(&A::i) << 'n';

 std::cout << "decltype(&B::i) == int B::* ? "
 << std::is_same<decltype(&B::i), int B::*>::value << 'n'; //#2
 B b;
 std::cout << b.*(&B::i) << 'n';

The code prints

decltype(&B::i) == int A::* ? true
10
decltype(&B::i) == int B::* ? false
10

I used the example in [expr.unary.op]/3, where the standard says that the type of &B::i is int A::*, but that is not normative.

c++ language-lawyer pointer-to-member

share|improve this question

edited 14 mins ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

add a comment | 

demo:

#include<iostream>
struct A int i = 10; ;
struct B : A ;

int main()
 std::cout << "decltype(&B::i) == int A::* ? " << std::boolalpha
 << std::is_same<decltype(&B::i), int A::*>::value << 'n'; //#1
 A a;
 std::cout << a.*(&A::i) << 'n';

 std::cout << "decltype(&B::i) == int B::* ? "
 << std::is_same<decltype(&B::i), int B::*>::value << 'n'; //#2
 B b;
 std::cout << b.*(&B::i) << 'n';

The code prints

decltype(&B::i) == int A::* ? true
10
decltype(&B::i) == int B::* ? false
10

I used the example in [expr.unary.op]/3, where the standard says that the type of &B::i is int A::*, but that is not normative.

c++ language-lawyer pointer-to-member

share|improve this question

edited 14 mins ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

#include<iostream>
struct A int i = 10; ;
struct B : A ;

int main()
 std::cout << "decltype(&B::i) == int A::* ? " << std::boolalpha
 << std::is_same<decltype(&B::i), int A::*>::value << 'n'; //#1
 A a;
 std::cout << a.*(&A::i) << 'n';

 std::cout << "decltype(&B::i) == int B::* ? "
 << std::is_same<decltype(&B::i), int B::*>::value << 'n'; //#2
 B b;
 std::cout << b.*(&B::i) << 'n';

decltype(&B::i) == int A::* ? true
10
decltype(&B::i) == int B::* ? false
10

share|improve this question

edited 14 mins ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

share|improve this question

edited 14 mins ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

edited 14 mins ago

Fabio Turati

2,75652542

edited 14 mins ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

asked 11 hours ago

AlexanderAlexander

985414

1 Answer
1

active

oldest

votes

7

From the paragraph you link to, emphasis mine:

If the operand is a qualified-id naming a non-static or variant member
m of some class C with type T, the result has type “pointer to member
of class C of type T” and is a prvalue designating C::m.

"Some class C" means it need not be the same class as the one mentioned by the qualified-id. In this case, i is a member of A, and remains a member of A even when named by &B::i. The type of &B::i is therefore int A::*, which you can verify by the test

std::is_same<decltype(&B::i), int A::*>::value

According to [class.qual]/1, member lookup follows the algorithm detailed in [class.member.lookup]. It is according to the rules there, which inspect the sub-object from which the member i comes from, that the class C is determined. Since i is a member of the sub-object A, the class of the pointer to member is determined to be A.

share|improve this answer

edited 10 hours ago

answered 10 hours ago

StoryTellerStoryTeller

112k17236303

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  To add some additional commentary: This follows on from the way that a pointer-to-member-of-base can be implicitly converted to pointer-to-member-of-derived (in contrast to the way that pointer-to-derived can be implicitly converted to pointer-to-base).
  
  – Martin Bonner
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not convinced by your conclusion: "Some class C means .... The type of &B::i is therefore int A::*.
  
  – Alexander
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Alexander - My conclusion is backed by the member lookup algorithm. And the wording of "some class" is not coincidental.
  
  – StoryTeller
  10 hours ago
  

  
  
  
  

add a comment | 

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7

From the paragraph you link to, emphasis mine:

If the operand is a qualified-id naming a non-static or variant member
m of some class C with type T, the result has type “pointer to member
of class C of type T” and is a prvalue designating C::m.

"Some class C" means it need not be the same class as the one mentioned by the qualified-id. In this case, i is a member of A, and remains a member of A even when named by &B::i. The type of &B::i is therefore int A::*, which you can verify by the test

std::is_same<decltype(&B::i), int A::*>::value

According to [class.qual]/1, member lookup follows the algorithm detailed in [class.member.lookup]. It is according to the rules there, which inspect the sub-object from which the member i comes from, that the class C is determined. Since i is a member of the sub-object A, the class of the pointer to member is determined to be A.

share|improve this answer

edited 10 hours ago

answered 10 hours ago

StoryTellerStoryTeller

112k17236303

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  To add some additional commentary: This follows on from the way that a pointer-to-member-of-base can be implicitly converted to pointer-to-member-of-derived (in contrast to the way that pointer-to-derived can be implicitly converted to pointer-to-base).
  
  – Martin Bonner
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not convinced by your conclusion: "Some class C means .... The type of &B::i is therefore int A::*.
  
  – Alexander
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Alexander - My conclusion is backed by the member lookup algorithm. And the wording of "some class" is not coincidental.
  
  – StoryTeller
  10 hours ago
  

  
  
  
  

add a comment | 

7

From the paragraph you link to, emphasis mine:

If the operand is a qualified-id naming a non-static or variant member
m of some class C with type T, the result has type “pointer to member
of class C of type T” and is a prvalue designating C::m.

"Some class C" means it need not be the same class as the one mentioned by the qualified-id. In this case, i is a member of A, and remains a member of A even when named by &B::i. The type of &B::i is therefore int A::*, which you can verify by the test

std::is_same<decltype(&B::i), int A::*>::value

According to [class.qual]/1, member lookup follows the algorithm detailed in [class.member.lookup]. It is according to the rules there, which inspect the sub-object from which the member i comes from, that the class C is determined. Since i is a member of the sub-object A, the class of the pointer to member is determined to be A.

share|improve this answer

edited 10 hours ago

answered 10 hours ago

StoryTellerStoryTeller

112k17236303

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  To add some additional commentary: This follows on from the way that a pointer-to-member-of-base can be implicitly converted to pointer-to-member-of-derived (in contrast to the way that pointer-to-derived can be implicitly converted to pointer-to-base).
  
  – Martin Bonner
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not convinced by your conclusion: "Some class C means .... The type of &B::i is therefore int A::*.
  
  – Alexander
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Alexander - My conclusion is backed by the member lookup algorithm. And the wording of "some class" is not coincidental.
  
  – StoryTeller
  10 hours ago
  

  
  
  
  

add a comment | 

From the paragraph you link to, emphasis mine:

If the operand is a qualified-id naming a non-static or variant member
m of some class C with type T, the result has type “pointer to member
of class C of type T” and is a prvalue designating C::m.

"Some class C" means it need not be the same class as the one mentioned by the qualified-id. In this case, i is a member of A, and remains a member of A even when named by &B::i. The type of &B::i is therefore int A::*, which you can verify by the test

std::is_same<decltype(&B::i), int A::*>::value

According to [class.qual]/1, member lookup follows the algorithm detailed in [class.member.lookup]. It is according to the rules there, which inspect the sub-object from which the member i comes from, that the class C is determined. Since i is a member of the sub-object A, the class of the pointer to member is determined to be A.

share|improve this answer

edited 10 hours ago

answered 10 hours ago

StoryTellerStoryTeller

112k17236303

std::is_same<decltype(&B::i), int A::*>::value

share|improve this answer

edited 10 hours ago

answered 10 hours ago

StoryTellerStoryTeller

112k17236303

answered 10 hours ago

StoryTellerStoryTeller

112k17236303

answered 10 hours ago

StoryTellerStoryTeller

112k17236303