CODEWITHSUNDEEP
c++scopenamespacesusing-directives
Rajdeep Sindhu
Rajdeep Sindhu

Reputation: 258

Unqualified name lookup after using-directives in C++

From cppreference :

From the point of view of unqualified name lookup of any name after a using-directive and until the end of the scope in which it appears, every name from namespace-name is visible as if it were declared in the nearest enclosing namespace which contains both the using-directive and namespace-name.

The "inner-most" scope that contains namespace-name is the scope in which namespace-name is declared, and it must enclose the scope containing the using-directive for the using-directive to be valid.

This implies that "the nearest enclosing namespace which contains both the using-directive and namespace-name" is just the namespace in which namespace-name has been declared. In that case, why phrase it in such a verbose manner?

Am I misunderstanding something? Are there any subtleties here that I'm missing?

Upvotes: 1

Views: 85

Answers (1)

user12002570
user12002570

Reputation: 1

is just the namespace in which namespace-name has been declared. In that case, why phrase it in such a verbose manner?

That is not always the case. To make it clear, consider the following contrived example. Here, the nearest enclosing namespace that contains both the using-directive and the namespace-name is the global namespace. This means that int k = i; is ambiguous because there are two i that can be used.

If on the other hand, if we were to follow your modified rule(just the namespace in which namespace-name has been declared), then the concerning namespace would be only outer that only contains the namespace-name.

int i = 12; 
namespace outer
{
    namespace n1
    {
        int i = 10;
    }
} 
namespace n2
{
    void f()
    {
        using namespace outer::n1; 
        int k = i;   //this is ambiguous because i already exists in global namespace
        
    } 
}

Upvotes: 2

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