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13

In C++ consider the grammar rule:

member-access-expression: LHS member-access-operator RHS

(op is .)

and
LHS=unqualified id-expression e.g. that references an instance variable.
RHS=qualified id-expression (with at least one nested identifier)

example: a.b::c

If that can ever pass the semantic check, what situation would it be ?

The following experiment:

struct B;

struct A

 B b;
;

int main()

 A a;
 a.b::c;

returns

'b' is not a class, namespace, or enumeration
a.b::c;
 ^

(demo)

This tends to hint to me that there can't be any legal case of a qualified-id on the right of a member access.

c++

share|improve this question

edited May 22 at 12:15

Lightness Races in Orbit

302k56489843

asked May 22 at 9:38

v.oddouv.oddou

4,27012449

add a comment | 

13

In C++ consider the grammar rule:

member-access-expression: LHS member-access-operator RHS

(op is .)

and
LHS=unqualified id-expression e.g. that references an instance variable.
RHS=qualified id-expression (with at least one nested identifier)

example: a.b::c

If that can ever pass the semantic check, what situation would it be ?

The following experiment:

struct B;

struct A

 B b;
;

int main()

 A a;
 a.b::c;

returns

'b' is not a class, namespace, or enumeration
a.b::c;
 ^

(demo)

This tends to hint to me that there can't be any legal case of a qualified-id on the right of a member access.

c++

share|improve this question

edited May 22 at 12:15

Lightness Races in Orbit

302k56489843

asked May 22 at 9:38

v.oddouv.oddou

4,27012449

add a comment | 

In C++ consider the grammar rule:

member-access-expression: LHS member-access-operator RHS

(op is .)

and
LHS=unqualified id-expression e.g. that references an instance variable.
RHS=qualified id-expression (with at least one nested identifier)

example: a.b::c

If that can ever pass the semantic check, what situation would it be ?

The following experiment:

struct B;

struct A

 B b;
;

int main()

 A a;
 a.b::c;

returns

'b' is not a class, namespace, or enumeration
a.b::c;
 ^

(demo)

This tends to hint to me that there can't be any legal case of a qualified-id on the right of a member access.

c++

share|improve this question

edited May 22 at 12:15

Lightness Races in Orbit

302k56489843

asked May 22 at 9:38

v.oddouv.oddou

4,27012449

struct B;

struct A

 B b;
;

int main()

 A a;
 a.b::c;

'b' is not a class, namespace, or enumeration
a.b::c;
 ^

share|improve this question

edited May 22 at 12:15

Lightness Races in Orbit

302k56489843

asked May 22 at 9:38

v.oddouv.oddou

4,27012449

share|improve this question

edited May 22 at 12:15

Lightness Races in Orbit

302k56489843

asked May 22 at 9:38

v.oddouv.oddou

4,27012449

edited May 22 at 12:15

Lightness Races in Orbit

302k56489843

edited May 22 at 12:15

Lightness Races in Orbit

302k56489843

asked May 22 at 9:38

v.oddouv.oddou

4,27012449

asked May 22 at 9:38

v.oddouv.oddou

4,27012449

3 Answers
3

active

oldest

votes

21

A very simple example is if you want to call a member function of a parent class:

struct A 
 void f();
;

struct B: A 
 void f();
;

B b;
b.A::f();

share|improve this answer

answered May 22 at 9:42

HoltHolt

26.8k65497

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  very interesting. then the "kind" referred-to by anything placed in the position of A has to be struct/class /union/namespace/enum ?
  
  – v.oddou
  May 22 at 9:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou It's not that surprising. Those "kinds" are what the scope resolution operator is used for.
  
  – Peter
  May 22 at 9:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Peter perfect. So I have everything: the grammar is valid; and the semantic situation where it's correct: when A refers to a symbol of kind user defined type.
  
  – v.oddou
  May 22 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With A looked up relatively from the scope of decltype(b) ? is it also legal to do b.::A::f() ? (seems to build fine in gcc)
  
  – v.oddou
  May 22 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  You don't even need inheritance. A::f() is a valid name for the function f() inside A, period. coliru.stacked-crooked.com/a/67d13365926585bd
  
  – Lightness Races in Orbit
  May 22 at 12:12
  
  

  
  
  
  

 | 
show 4 more comments

9

A use case is accessing members of an enum within some struct A by using an instance of A (rather than using the enum directly via A::b::c):

struct A 
 enum class b c ; // can be unscoped as well
;

A a;
a.b::c; // Access to enum value c - similarly, A::b::c would work

share|improve this answer

edited May 22 at 20:47

answered May 22 at 11:13

andreeeandreee

2,1261125

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  mesmerizing. I knew you could use an instance to access a static field, but not a type. Your answer is important for me to widen the exhaustivity of the use cases of id-expression in combination with member-access-expression. thanks !
  
  – v.oddou
  May 23 at 2:52
  

  
  
  
  

add a comment | 

5

Here's a trivial example:

struct A 
 void f() 
;

int main()

 A a;
 a.A::f();

A::f() is a qualified version of the name for the function f that's a member of A. You can use it in member access just like the "short" (or unqualified) name.

In fact, one might argue that every time you write a.f(), that's a shortcut for a.A::f() (with the A:: part being taken automatically from decltype(a)).

There's nothing magic about this, though it's unusual to see the construct outside of the sort of scenarios the other answerers demonstrated, because in this example it's redundant.

share|improve this answer

edited May 23 at 10:33

answered May 22 at 12:13

Lightness Races in OrbitLightness Races in Orbit

302k56489843

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  superb. This is a 3rd example of different flavor which brings value. It confirms that the RHS of a member-access is looked up from the scope of its LHS but the result of the evaluation of the LHS sub-expression is not necessarily used.
  
  – v.oddou
  May 23 at 2:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  fully-qualified is it ? if it lacks the global scope it is qualified but not fully, IMHO.
  
  – v.oddou
  May 23 at 2:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou Hmm... good question! I don't know actually. It's not a standard term anyway and I can't remember what conventional usage is. I'll just rip out the "fully-" as it's a distraction anyway.
  
  – Lightness Races in Orbit
  May 23 at 10:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou Looks like you're right!
  
  – Lightness Races in Orbit
  May 23 at 10:46
  

  
  
  
  

add a comment | 

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21

A very simple example is if you want to call a member function of a parent class:

struct A 
 void f();
;

struct B: A 
 void f();
;

B b;
b.A::f();

share|improve this answer

answered May 22 at 9:42

HoltHolt

26.8k65497

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  very interesting. then the "kind" referred-to by anything placed in the position of A has to be struct/class /union/namespace/enum ?
  
  – v.oddou
  May 22 at 9:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou It's not that surprising. Those "kinds" are what the scope resolution operator is used for.
  
  – Peter
  May 22 at 9:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Peter perfect. So I have everything: the grammar is valid; and the semantic situation where it's correct: when A refers to a symbol of kind user defined type.
  
  – v.oddou
  May 22 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With A looked up relatively from the scope of decltype(b) ? is it also legal to do b.::A::f() ? (seems to build fine in gcc)
  
  – v.oddou
  May 22 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  You don't even need inheritance. A::f() is a valid name for the function f() inside A, period. coliru.stacked-crooked.com/a/67d13365926585bd
  
  – Lightness Races in Orbit
  May 22 at 12:12
  
  

  
  
  
  

 | 
show 4 more comments

21

A very simple example is if you want to call a member function of a parent class:

struct A 
 void f();
;

struct B: A 
 void f();
;

B b;
b.A::f();

share|improve this answer

answered May 22 at 9:42

HoltHolt

26.8k65497

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  very interesting. then the "kind" referred-to by anything placed in the position of A has to be struct/class /union/namespace/enum ?
  
  – v.oddou
  May 22 at 9:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou It's not that surprising. Those "kinds" are what the scope resolution operator is used for.
  
  – Peter
  May 22 at 9:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Peter perfect. So I have everything: the grammar is valid; and the semantic situation where it's correct: when A refers to a symbol of kind user defined type.
  
  – v.oddou
  May 22 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  With A looked up relatively from the scope of decltype(b) ? is it also legal to do b.::A::f() ? (seems to build fine in gcc)
  
  – v.oddou
  May 22 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  You don't even need inheritance. A::f() is a valid name for the function f() inside A, period. coliru.stacked-crooked.com/a/67d13365926585bd
  
  – Lightness Races in Orbit
  May 22 at 12:12
  
  

  
  
  
  

 | 
show 4 more comments

A very simple example is if you want to call a member function of a parent class:

struct A 
 void f();
;

struct B: A 
 void f();
;

B b;
b.A::f();

share|improve this answer

answered May 22 at 9:42

HoltHolt

26.8k65497

struct A 
 void f();
;

struct B: A 
 void f();
;

B b;
b.A::f();

share|improve this answer

answered May 22 at 9:42

HoltHolt

26.8k65497

answered May 22 at 9:42

HoltHolt

26.8k65497

answered May 22 at 9:42

HoltHolt

26.8k65497

9

A use case is accessing members of an enum within some struct A by using an instance of A (rather than using the enum directly via A::b::c):

struct A 
 enum class b c ; // can be unscoped as well
;

A a;
a.b::c; // Access to enum value c - similarly, A::b::c would work

share|improve this answer

edited May 22 at 20:47

answered May 22 at 11:13

andreeeandreee

2,1261125

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  mesmerizing. I knew you could use an instance to access a static field, but not a type. Your answer is important for me to widen the exhaustivity of the use cases of id-expression in combination with member-access-expression. thanks !
  
  – v.oddou
  May 23 at 2:52
  

  
  
  
  

add a comment | 

9

A use case is accessing members of an enum within some struct A by using an instance of A (rather than using the enum directly via A::b::c):

struct A 
 enum class b c ; // can be unscoped as well
;

A a;
a.b::c; // Access to enum value c - similarly, A::b::c would work

share|improve this answer

edited May 22 at 20:47

answered May 22 at 11:13

andreeeandreee

2,1261125

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  mesmerizing. I knew you could use an instance to access a static field, but not a type. Your answer is important for me to widen the exhaustivity of the use cases of id-expression in combination with member-access-expression. thanks !
  
  – v.oddou
  May 23 at 2:52
  

  
  
  
  

add a comment | 

A use case is accessing members of an enum within some struct A by using an instance of A (rather than using the enum directly via A::b::c):

struct A 
 enum class b c ; // can be unscoped as well
;

A a;
a.b::c; // Access to enum value c - similarly, A::b::c would work

share|improve this answer

edited May 22 at 20:47

answered May 22 at 11:13

andreeeandreee

2,1261125

struct A 
 enum class b c ; // can be unscoped as well
;

A a;
a.b::c; // Access to enum value c - similarly, A::b::c would work

share|improve this answer

edited May 22 at 20:47

answered May 22 at 11:13

andreeeandreee

2,1261125

answered May 22 at 11:13

andreeeandreee

2,1261125

answered May 22 at 11:13

andreeeandreee

2,1261125

5

Here's a trivial example:

struct A 
 void f() 
;

int main()

 A a;
 a.A::f();

A::f() is a qualified version of the name for the function f that's a member of A. You can use it in member access just like the "short" (or unqualified) name.

In fact, one might argue that every time you write a.f(), that's a shortcut for a.A::f() (with the A:: part being taken automatically from decltype(a)).

There's nothing magic about this, though it's unusual to see the construct outside of the sort of scenarios the other answerers demonstrated, because in this example it's redundant.

share|improve this answer

edited May 23 at 10:33

answered May 22 at 12:13

Lightness Races in OrbitLightness Races in Orbit

302k56489843

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  superb. This is a 3rd example of different flavor which brings value. It confirms that the RHS of a member-access is looked up from the scope of its LHS but the result of the evaluation of the LHS sub-expression is not necessarily used.
  
  – v.oddou
  May 23 at 2:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  fully-qualified is it ? if it lacks the global scope it is qualified but not fully, IMHO.
  
  – v.oddou
  May 23 at 2:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou Hmm... good question! I don't know actually. It's not a standard term anyway and I can't remember what conventional usage is. I'll just rip out the "fully-" as it's a distraction anyway.
  
  – Lightness Races in Orbit
  May 23 at 10:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou Looks like you're right!
  
  – Lightness Races in Orbit
  May 23 at 10:46
  

  
  
  
  

add a comment | 

5

Here's a trivial example:

struct A 
 void f() 
;

int main()

 A a;
 a.A::f();

A::f() is a qualified version of the name for the function f that's a member of A. You can use it in member access just like the "short" (or unqualified) name.

In fact, one might argue that every time you write a.f(), that's a shortcut for a.A::f() (with the A:: part being taken automatically from decltype(a)).

There's nothing magic about this, though it's unusual to see the construct outside of the sort of scenarios the other answerers demonstrated, because in this example it's redundant.

share|improve this answer

edited May 23 at 10:33

answered May 22 at 12:13

Lightness Races in OrbitLightness Races in Orbit

302k56489843

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  superb. This is a 3rd example of different flavor which brings value. It confirms that the RHS of a member-access is looked up from the scope of its LHS but the result of the evaluation of the LHS sub-expression is not necessarily used.
  
  – v.oddou
  May 23 at 2:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  fully-qualified is it ? if it lacks the global scope it is qualified but not fully, IMHO.
  
  – v.oddou
  May 23 at 2:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou Hmm... good question! I don't know actually. It's not a standard term anyway and I can't remember what conventional usage is. I'll just rip out the "fully-" as it's a distraction anyway.
  
  – Lightness Races in Orbit
  May 23 at 10:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @v.oddou Looks like you're right!
  
  – Lightness Races in Orbit
  May 23 at 10:46
  

  
  
  
  

add a comment | 

Here's a trivial example:

struct A 
 void f() 
;

int main()

 A a;
 a.A::f();

A::f() is a qualified version of the name for the function f that's a member of A. You can use it in member access just like the "short" (or unqualified) name.

In fact, one might argue that every time you write a.f(), that's a shortcut for a.A::f() (with the A:: part being taken automatically from decltype(a)).

There's nothing magic about this, though it's unusual to see the construct outside of the sort of scenarios the other answerers demonstrated, because in this example it's redundant.

share|improve this answer

edited May 23 at 10:33

answered May 22 at 12:13

Lightness Races in OrbitLightness Races in Orbit

302k56489843

struct A 
 void f() 
;

int main()

 A a;
 a.A::f();

share|improve this answer

edited May 23 at 10:33

answered May 22 at 12:13

Lightness Races in OrbitLightness Races in Orbit

302k56489843

answered May 22 at 12:13

Lightness Races in OrbitLightness Races in Orbit

302k56489843

answered May 22 at 12:13

Lightness Races in OrbitLightness Races in Orbit

302k56489843