Access outer struct with the same name

Question

Look at my sample code

struct A
{
   int member;
};

int main()
{
   int A; //Line 1
   A b;   //Line 2 
   b.member = int(); //Line 3
}

Errors are

prog.cpp: In function ‘int main()’:
prog.cpp:9: error: expected `;' before ‘b’
prog.cpp:9: warning: statement has no effect
prog.cpp:10: error: ‘b’ was not declared in this scope

How to access structure A in second line ? Why do I get the error anyway?

Answer 1

How to remove the error in Line 2?

Use Elaborated Type Specifier, ie instead of writing A b; write struct A b; .

3.4.4 Elaborated type specifiers

An elaborated-type-specifier may be used to refer to a previously declared class-name or enum-name even though the name has been hidden by a non-type declaration (3.3.7) . The class-name or enum-name in the elaborated-type-specifier may either be a simple identifer or be a qualified-id.

---

Why do I get the error anyway?

Because A outside main is hidden inside main after the definition of int A . The only way to access struct A is by using elaborated-type-specifier.

3.3.7 Name hiding

2) A class name (9.1) or enumeration name (7.2) can be hidden by the name of an object, function, or enumerator declared in the same scope. If a class or enumeration name and an object, function, or enumerator are declared in the same scope (in any order) with the same name , the class or enumeration name is hidden wherever the object, function, or enumerator name is visible.