Unqualified name lookup after using-directives in C++

Question

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?

"in which namespace-name has been defined" I suppose you mean "declared" not "defined" — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented Jun 29, 2024 at 10:34
The same (enclosing) namespace can be opened and closed more than once in the same translation unit. It's being specific about which one the using applies to. — Richard Critten
– Richard Critten, Commented Jun 29, 2024 at 10:41
@RichardCritten using applies to the entire merged namespace in that case, doesn't it? — Rajdeep Sindhu
– Rajdeep Sindhu, Commented Jun 29, 2024 at 10:51
Your title talks about using-declarations but your question about using-directives. They are two distinct language constructs and have completely different effects on name lookup. A using-directive is of the form using namespace X;. A using-declaration is of the form using X::Y;. (To make it more confusing both using-directives and using-declarations are declarations.) — user17732522
– user17732522, Commented Jun 29, 2024 at 16:02
@user17732522 Fixed it, my bad. I got the tag right but messed up the title. — Rajdeep Sindhu
– Rajdeep Sindhu, Commented Jun 29, 2024 at 16:50

Richard · Accepted Answer · 2024-06-29 11:32:59Z

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 } }

Ah, I was wrong in assuming that the scope in which namespace-name is declared must have the using-directive.

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