问题
There are 3 examples:
I.
typedef int foo;
namespace B
{
struct S
{
operator int(){ return 24; }
};
int foo(B::S s){ return 0; }
}
int main()
{
int t=foo(B::S()); //24, ADL does not apply
}
II.
namespace B
{
struct S
{
operator int(){ return 24; }
};
int foo(B::S s){ return 0; }
}
int main()
{
int t=foo(B::S()); //0, ADL applies
}
III.
namespace B
{
struct S
{
operator int(){ return 24; }
};
int foo(B::S s){ return 0; }
}
int foo(B::S s){ return 12; }
int main()
{
int t=foo(B::S()); //error: call of overloaded ‘foo(B::S)’ is ambiguous
//ADL applies
}
It is not clear for me what is the actual conditions to ADL lookup will be apply? I need in reference to standard described it.
回答1:
This Standard paragraph clarifies, and even has an example very much like your first example.
3.4.1/3:
The lookup for an unqualified name used as the postfix-expression of a function call is described in 3.4.2 [basic.lookup.argdep]. [Note: For purposes of determining (during parsing) whether an expression is a postfix-expression for a function call, the usual name lookup rules apply. The rules in 3.4.2 have no effect on the syntactic interpretation of an expression. For example,
typedef int f;
namespace N {
struct A {
friend void f(A &);
operator int();
void g(A a) {
int i = f(a); // f is the typedef, not the friend
// function: equivalent to int(a)
}
};
}
Because the expression is not a function call, the argument-dependent name lookup (3.4.2) does not apply and the friend function
f
is not found. -end note]
回答2:
Your first example does not illustrate ADL. In the line
int t=foo(B::S());
foo
is typedef
ed to int
.
The following code has some better illustrations of ADL.
#include <iostream>
namespace B
{
struct S
{
operator int(){ return 24; }
};
int foo(S s){ return 100; }
int bar(S s){ return 400; }
}
namespace C
{
struct S
{
operator int(){ return 24; }
};
int foo(S s){ return 200; }
}
int bar(C::S s){ return 800; }
int main()
{
// ADL makes it possible for foo to be resolved to B::foo
std::cout << foo(B::S()) << std::endl;
// ADL makes it possible for foo to be resolved to C::foo
std::cout << foo(C::S()) << std::endl;
// ADL makes it possible for bar to be resolved to B::bar
std::cout << bar(B::S()) << std::endl;
// ADL makes it possible for bar to be resolved to ::bar
std::cout << bar(C::S()) << std::endl;
}
来源:https://stackoverflow.com/questions/23958590/when-is-adl-applied