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Question #126 Difficulty: 2

According to the C++23 standard, what is the output of this program? 

    #include<iostream>

int foo()
{
  return 10;
}

struct foobar
{
  static int x;
  static int foo()
  {
    return 11;
  }
};

int foobar::x = foo();

int main()
{
    std::cout << foobar::x;
}

Correct!

Will the global foo or the member foo be chosen?

First, notice that the name foo is unqualified. §[basic.lookup.unqual]¶4:

An unqualified name is a name that does not immediately follow a nested-name-specifier or the . or -> in a class member access expression (§[expr.ref]), possibly after a template keyword or ~. Unless otherwise specified, such a name undergoes unqualified name lookup from the point where it appears.

What does unqualified name lookup mean? §[basic.lookup.unqual]¶3:

Unqualified name lookup from a program point performs an unqualified search in its immediate scope.

The definition of an unqualified search is given in §[basic.lookup.unqual]¶2:

An unqualified search in a scope S from a program point P includes the results of searches from P in

— S, and

— for any scope U that contains P and is or is contained by S, each namespace contained by S that is nominated by a using-directive that is active in U at P.

If no declarations are found, the results of the unqualified search are the results of an unqualified search in the parent scope of S, if any, from P.

So, since we don't have any using-directives, we search for foo in its immediate scope and if we find any result, we stop searching. (If we don't, we keep searching in parent scopes, recursively.)

The immediate scope at some program point is the smallest scope that contains it (§[basic.scope.scope]¶1). What is the immediate scope at the program point where we call foo? §[basic.scope.class]¶1:

Any declaration of a class or class template C introduces a class scope that includes the member-specification of the class-specifier for C (if any). For each non-friend redeclaration or specialization whose target scope is or is contained by the scope, the portion after the declarator-id, class-head-name, or enum-head-name is also included in the scope.

In the statement int foobar::x = foo(); we redeclare the variable x. Its target scope is a class scope introduced by the declaration of the class foobar (§[dcl.meaning.general]¶3.4). The portion after x is also included in that scope. So when looking for foo we find foobar::foo there and don't search any further.

You can explore this question further on C++ Insights or Compiler Explorer!

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