Under what circumstances can an extern variable be used in definition?

Question

I am very very sorry. I didn\'t know my incomplete code attachment would create such a mess. I am very glad to see so many sincere helps.

This code will compile:

Answer 1

The following is a declaration and definition:

int x;

Adding extern says "make it a declaration only, please".

But when you are providing a value, the line has to be a definition, so the variable gets extern storage class and you just happen to be defining it right in place anyway:

extern int x = 3;

The linkage semantics are as they usually are for extern, and the storage location is just as it would be for a normal definition int x = 3 — i.e. in that TU at namespace scope. myadd is not relevant at all.

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It's a hard one to "prove", because it's a case of "there's no rule against it".

Here's the best quote:

[n3290: 3.1/2]: A declaration is a definition unless it declares a function without specifying the function’s body (8.4), it contains the extern specifier (7.1.1) or a linkage-specification25 (7.5) and neither an initializer nor a function-body, [..]

And some other pertinent information:

[n3290: 3.5/2]: A name is said to have linkage when it might denote the same object, reference, function, type, template, namespace or value as a name introduced by a declaration in another scope:

• When a name has external linkage, the entity it denotes can be referred to by names from scopes of other translation units or from other scopes of the same translation unit.
• When a name has internal linkage, the entity it denotes can be referred to by names from other scopes in the same translation unit.
• When a name has no linkage, the entity it denotes cannot be referred to by names from other scopes.

[n3290: 3.5/12]:The name of a function declared in block scope and the name of a variable declared by a block scope extern declaration have linkage. If there is a visible declaration of an entity with linkage having the same name and type, ignoring entities declared outside the innermost enclosing namespace scope, the block scope declaration declares that same entity and receives the linkage of the previous declaration. If there is more than one such matching entity, the program is ill-formed. Otherwise, if no matching entity is found, the block scope entity receives external linkage. [..]