c++ - Why does the compiler stops the name lookup on overloads? -

c++ - Why does the compiler stops the name lookup on overloads? -

i read article:fun c++ namespaces author shows compiler stops looking overloads when encountering first one, here using namespaces.

namespace { void f(int x); // our std::sqrt(double) } namespace b { struct s {}; // user-defined type associated namespace b void f(s); void f(int, int); void test1() { using namespace a; // using directive f(1); // error namespace not considered because // b contains 2 overloads 'f' f(1,2); // ok b::f(int,int) f(b::s()); // ok b::f(s) } void test2() { using a::f; // using declaration f(1); // ok a::f(int) f(1,2); // error a::f hides b::f(int,int) f(b::s()); // ok b::f(s) due adl! } } namespace c { void test3() { using namespace a; // using directive f(1); // ok a::f(int) f(b::s()); // ok b::f(s) due adl! } void test4() { using a::f; // using declaration f(1); // ok a::f(int) f(b::s()); // ok b::f(s) due adl! } }

why compiler supposed stop?

edit #1: question indeed ment be: why standard says so?

thanks answers!

the compiler stops looking overloads when encountering first one

no, doesn't stop "when encountering first one" otherwise couldn't find both b::f(int,int) , b::f(s).

it finds overloads in given scope (not first one), doesn't farther in more distant scopes.

that's name lookup in c++, if have global variable called var , in function have local variable called var, using name within function refer local variable. it's more useful way, it's more meant utilize variable in declared nearby, it's in related code.

if hands letter , tells give fred, standing few metres away wearing badge says "i fred", ignore him , go outside , maintain looking every other person in world called fred?

c++ compiler-construction