Where in the Standard (C++14) does it say that the following two declarations are equivalent?

Question

struct A{};
int A;
struct A a;
struct A::A b;

The last two declarations above are equivalent.They both declare objects of type A. Where in the Standard can I find or deduce this?

Answer 1

[class]/2:

A class-name is inserted into the scope in which it is declared immediately after the class-name is seen. The class-name is also inserted into the scope of the class itself; this is known as the injected-class-name.

I.e. A::A::A::A refers to A as well. In some contexts, A::A could name the constructor instead, though - [class.qual]/2 covers this, and its note even addresses your example:

In a lookup in which function names are not ignored³³ and the nested-name-specifier nominates a class C

• if the name specified after the nested-name-specifier, when looked up in C, is the injected-class-name of C (Clause 9), or
• in a using-declaration (7.3.3) that is a member-declaration, if the name specified after the nested-name- specifier is the same as the identifier or the simple-template-id’s template-name in the last component of the nested-name-specifier,

the name is instead considered to name the constructor of class C. [ Note: For example, the constructor is not an acceptable lookup result in an elaborated-type-specifier so the constructor would not be used in place of the injected-class-name. — end note ]

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³³⁾ Lookups in which function names are ignored include names appearing in a nested-name-specifier, an elaborated-type- specifier, or a base-specifier.

So in a statement such as

A::A a;

Function names are not ignored when looking up A::A, and thus the code is ill-formed as A::A refers to the constructor. However, in

struct B : A::A {};
struct A::A a;

Everything is fine as function names are ignored in base-specifiers and elaborated-type-specifiers.