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How would you explain this difference in pointer to members of base and derived class using standard quotes?

How to call a parent class function from derived class function?Pointer to class data member “::*”Purpose of Unions in C and C++Pretty-print C++ STL containersEfficient unsigned-to-signed cast avoiding implementation-defined behaviorWhy is f(i = -1, i = -1) undefined behavior?Why does the C++ linker allow undefined functions?restrict a template function, to only allow certain typesConverting a pointer-to-member-of-base to a pointer-to-member-of-derivedDoes the C++ standard allow for an uninitialized bool to crash a program?

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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