What happens when a class and a function have the same name?

Question

#include <iostream>
using namespace std;

struct test
{
    test(){cout<<"class"<<endl;}
};
void test(){cout<<"function"<<endl;}

int main()
{
    test();
    return 0;
}

Output:

function  

(VS2013 ang gcc 4.8.1)

Why function is selected? Isn't it ambiguity?

Answer 1

This is called name hiding and described in

3.3 Scope [basic.scope]

3.3.1 Declarative regions and scopes [basic.scope.declarative]

4) Given a set of declarations in a single declarative region, each of which specifies the same unqualified name,
— they shall all refer to the same entity, or all refer to functions and function templates; or
— exactly one declaration shall declare a class name or enumeration name that is not a typedef name and the other declarations shall all refer to the same variable or enumerator, or all refer to functions and function templates; in this case the class name or enumeration name is hidden (3.3.10). [...]

emphasis mine.

Note that changing the order of declaration doesn't affect the outcome:

void test(){cout<<"function"<<endl;}

struct test
{
    test(){cout<<"class"<<endl;}
};

int main()
{
    test();
    return 0;
}

still prints out function.

In case it isn't obvious, don't do this :)

Answer 2

From N3485 §3.3.10 [basic.scope.hiding]/2:

A class name (9.1) or enumeration name (7.2) can be hidden by the name of a variable, data member, function, or enumerator declared in the same scope.

Therefore, the function takes precedence over the class.

As mentioned in the comments, the class is still accessible via the class or struct keyword. If the class took precedence, the function would be unreachable.