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A fortified nonogram

What is a fully qualified name?

Is there a context where the expression `a.b::c` makes sense?What is the difference between #include <filename> and #include “filename”?What are the differences between a pointer variable and a reference variable in C++?What does the explicit keyword mean?What is the “-->” operator in C++?What is the copy-and-swap idiom?What is The Rule of Three?What are the basic rules and idioms for operator overloading?C++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming?Image Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionWhat are the new features in C++17?

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13

As far as I can tell, the term fully qualified isn't mentioned in the standard (e.g.), but I can recall "hearing" it many times online.

What do people mean when they say a name is fully qualified?

Does this count?

A::f()

or only this?

::A::f()

And, if it is standard, which wording have I not found?

c++ language-lawyer c++17

share|improve this question

edited 7 hours ago

dfri

36.9k462104

asked 8 hours ago

Lightness Races in OrbitLightness Races in Orbit

300k56485833

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  With the injected class name one can really qualify stuff. ::A::A::A::A::f(). Not sure if it's "fully" though.
  
  – StoryTeller
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @StoryTeller We should petition to add that term to the standard! really-really-really-qualified-id etc
  
  – Lightness Races in Orbit
  7 hours ago
  

  
  
  
  

add a comment | 

13

As far as I can tell, the term fully qualified isn't mentioned in the standard (e.g.), but I can recall "hearing" it many times online.

What do people mean when they say a name is fully qualified?

Does this count?

A::f()

or only this?

::A::f()

And, if it is standard, which wording have I not found?

c++ language-lawyer c++17

share|improve this question

edited 7 hours ago

dfri

36.9k462104

asked 8 hours ago

Lightness Races in OrbitLightness Races in Orbit

300k56485833

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  With the injected class name one can really qualify stuff. ::A::A::A::A::f(). Not sure if it's "fully" though.
  
  – StoryTeller
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @StoryTeller We should petition to add that term to the standard! really-really-really-qualified-id etc
  
  – Lightness Races in Orbit
  7 hours ago
  

  
  
  
  

add a comment | 

As far as I can tell, the term fully qualified isn't mentioned in the standard (e.g.), but I can recall "hearing" it many times online.

What do people mean when they say a name is fully qualified?

Does this count?

A::f()

or only this?

::A::f()

And, if it is standard, which wording have I not found?

c++ language-lawyer c++17

share|improve this question

edited 7 hours ago

dfri

36.9k462104

asked 8 hours ago

Lightness Races in OrbitLightness Races in Orbit

300k56485833

A::f()

::A::f()

share|improve this question

edited 7 hours ago

dfri

36.9k462104

asked 8 hours ago

Lightness Races in OrbitLightness Races in Orbit

300k56485833

share|improve this question

edited 7 hours ago

dfri

36.9k462104

asked 8 hours ago

Lightness Races in OrbitLightness Races in Orbit

300k56485833

edited 7 hours ago

dfri

36.9k462104

edited 7 hours ago

dfri

36.9k462104

asked 8 hours ago

Lightness Races in OrbitLightness Races in Orbit

300k56485833

asked 8 hours ago

Lightness Races in OrbitLightness Races in Orbit

300k56485833

2 Answers
2

active

oldest

votes

15

An identifier that uses the scope resolution operator is a qualified name as per [expr.prim.id.qual]. Otherwise it is unqualified.

The standard doesn't define the meaning of fully qualified, but it does mention it in [library]/[requirements]/[organization]/[contents] which says

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.
For example, if the Effects: element for library function F is described as calling library function G, the function ::std::G is meant.

Wikipedia defines Fully qualified name:

In computer programming, a fully qualified name is an unambiguous name that specifies which object, function, or variable a call refers to without regard to the context of the call

Only a name qualified starting from the global namespace is unambiguous without context. This is the common usage.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

eerorikaeerorika

92.7k668139

add a comment | 

12

Indeed, it is not a standard term. It has no definition in the standard.

However, the phrase "fully qualified" appears exactly once, in [contents] (15.5.1.1 "Library contents" in the as-of-writing most current draft N4800) paragraph 3:

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.

So in this definition, only names starting with :: are fully qualified.

share|improve this answer

answered 8 hours ago

Sebastian RedlSebastian Redl

51.7k478121

add a comment | 

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15

An identifier that uses the scope resolution operator is a qualified name as per [expr.prim.id.qual]. Otherwise it is unqualified.

The standard doesn't define the meaning of fully qualified, but it does mention it in [library]/[requirements]/[organization]/[contents] which says

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.
For example, if the Effects: element for library function F is described as calling library function G, the function ::std::G is meant.

Wikipedia defines Fully qualified name:

In computer programming, a fully qualified name is an unambiguous name that specifies which object, function, or variable a call refers to without regard to the context of the call

Only a name qualified starting from the global namespace is unambiguous without context. This is the common usage.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

eerorikaeerorika

92.7k668139

add a comment | 

15

An identifier that uses the scope resolution operator is a qualified name as per [expr.prim.id.qual]. Otherwise it is unqualified.

The standard doesn't define the meaning of fully qualified, but it does mention it in [library]/[requirements]/[organization]/[contents] which says

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.
For example, if the Effects: element for library function F is described as calling library function G, the function ::std::G is meant.

Wikipedia defines Fully qualified name:

In computer programming, a fully qualified name is an unambiguous name that specifies which object, function, or variable a call refers to without regard to the context of the call

Only a name qualified starting from the global namespace is unambiguous without context. This is the common usage.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

eerorikaeerorika

92.7k668139

add a comment | 

An identifier that uses the scope resolution operator is a qualified name as per [expr.prim.id.qual]. Otherwise it is unqualified.

The standard doesn't define the meaning of fully qualified, but it does mention it in [library]/[requirements]/[organization]/[contents] which says

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.
For example, if the Effects: element for library function F is described as calling library function G, the function ::std::G is meant.

Wikipedia defines Fully qualified name:

In computer programming, a fully qualified name is an unambiguous name that specifies which object, function, or variable a call refers to without regard to the context of the call

Only a name qualified starting from the global namespace is unambiguous without context. This is the common usage.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

eerorikaeerorika

92.7k668139

share|improve this answer

edited 8 hours ago

answered 8 hours ago

eerorikaeerorika

92.7k668139

answered 8 hours ago

eerorikaeerorika

92.7k668139

answered 8 hours ago

eerorikaeerorika

92.7k668139

12

Indeed, it is not a standard term. It has no definition in the standard.

However, the phrase "fully qualified" appears exactly once, in [contents] (15.5.1.1 "Library contents" in the as-of-writing most current draft N4800) paragraph 3:

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.

So in this definition, only names starting with :: are fully qualified.

share|improve this answer

answered 8 hours ago

Sebastian RedlSebastian Redl

51.7k478121

add a comment | 

12

Indeed, it is not a standard term. It has no definition in the standard.

However, the phrase "fully qualified" appears exactly once, in [contents] (15.5.1.1 "Library contents" in the as-of-writing most current draft N4800) paragraph 3:

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.

So in this definition, only names starting with :: are fully qualified.

share|improve this answer

answered 8 hours ago

Sebastian RedlSebastian Redl

51.7k478121

add a comment | 

Indeed, it is not a standard term. It has no definition in the standard.

However, the phrase "fully qualified" appears exactly once, in [contents] (15.5.1.1 "Library contents" in the as-of-writing most current draft N4800) paragraph 3:

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise.

So in this definition, only names starting with :: are fully qualified.

share|improve this answer

answered 8 hours ago

Sebastian RedlSebastian Redl

51.7k478121

share|improve this answer

answered 8 hours ago

Sebastian RedlSebastian Redl

51.7k478121

answered 8 hours ago

Sebastian RedlSebastian Redl

51.7k478121

answered 8 hours ago

Sebastian RedlSebastian Redl

51.7k478121