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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 1 hour 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 1 hour 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 1 hour 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 1 hour ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

share|improve this question

edited 1 hour ago

Fabio Turati

2,75652542

asked 11 hours ago

AlexanderAlexander

985414

edited 1 hour ago

Fabio Turati

2,75652542

edited 1 hour 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 11 hours ago

answered 11 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
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Alexander - My conclusion is backed by the member lookup algorithm. And the wording of "some class" is not coincidental.
  
  – StoryTeller
  11 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 11 hours ago

answered 11 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
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Alexander - My conclusion is backed by the member lookup algorithm. And the wording of "some class" is not coincidental.
  
  – StoryTeller
  11 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 11 hours ago

answered 11 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
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Alexander - My conclusion is backed by the member lookup algorithm. And the wording of "some class" is not coincidental.
  
  – StoryTeller
  11 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 11 hours ago

answered 11 hours ago

StoryTellerStoryTeller

112k17236303

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

share|improve this answer

edited 11 hours ago

answered 11 hours ago

StoryTellerStoryTeller

112k17236303

answered 11 hours ago

StoryTellerStoryTeller

112k17236303

answered 11 hours ago

StoryTellerStoryTeller

112k17236303