Qualified name lookup

#include <iostream>
 
namespace M {
    const char* fail = "fail\n";
}
 
using M::fail;
 
namespace N {
    const char* ok = "ok\n";
}
 
using namespace N;
 
int main()
{
    struct std {};
 
    std::cout << ::fail; // Error: unqualified lookup for 'std' finds the struct
    ::std::cout << ::ok; // OK: ::std finds the namespace std
}

struct A
{
    static int n;
};
 
int main()
{
    int A;
    A::n = 42; // OK: unqualified lookup of A to the left of :: ignores the variable
    A b;       // Error: unqualified lookup of A finds the variable A
}
 
template<int>
struct B : A {};
 
namespace N
{
    template<int>
    void B();
 
    int f()
    {
        return B<0>::n; // Error: N::B<0> is not a type
    }
}

class X {};
 
constexpr int number = 100;
 
struct C
{
    class X {};
    static const int number = 50;
    static X arr[number];
};
 
X C::arr[number], brr[number];    // Error: look up for X finds ::X, not C::X
C::X C::arr[number], brr[number]; // OK: size of arr is 50, size of brr is 100

struct C { typedef int I; };
 
typedef int I1, I2;
 
extern int *p, *q;
 
struct A { ~A(); };
 
typedef A AB;
 
int main()
{
    p->C::I::~I(); // The name I after ~ is looked up in the same scope as I before ::
                   // (that is, within the scope of C, so it finds C::I)
 
    q->I1::~I2();  // The name I2 is looked up in the same scope as I1
                   // (that is, from the current scope, so it finds ::I2)
 
    AB x;
    x.AB::~AB();   // The name AB after ~ is looked up in the same scope as AB before ::
                   // (that is, from the current scope, so it finds ::AB)
}

struct A { A(); };
 
struct B : A { B(); };
 
A::A() {} // A::A names a constructor, used in a declaration
B::B() {} // B::B names a constructor, used in a declaration
 
B::A ba;  // B::A names the type A (looked up in the scope of B)
A::A a;   // Error: A::A does not name a type
 
struct A::A a2; // OK: lookup in elaborated type specifier ignores functions
                // so A::A simply names the class A as seen from within the scope of A
                // (that is, the injected-class-name)

struct B { virtual void foo(); };
 
struct D : B { void foo() override; };
 
int main()
{
    D x;
    B& b = x;
 
    b.foo();    // Calls D::foo (virtual dispatch)
    b.B::foo(); // Calls B::foo (static dispatch)
}

namespace N
{
    template<typename T>
    struct foo {};
 
    struct X {};
}
 
N::foo<X> x; // Error: X is looked up as ::X, not as N::X

int x;
 
namespace Y
{
    void f(float);
    void h(int);
}
 
namespace Z
{
    void h(double);
}
 
namespace A
{
    using namespace Y;
    void f(int);
    void g(int);
    int i;
}
 
namespace B
{
    using namespace Z;
    void f(char);
    int i;
}
 
namespace AB
{
    using namespace A;
    using namespace B;
    void g();
}
 
void h()
{
    AB::g();  // AB is searched, AB::g found by lookup and is chosen AB::g(void)
              // (A and B are not searched)
 
    AB::f(1); // First, AB is searched. There is no f
              // Then, A, B are searched
              // A::f, B::f found by lookup
              // (but Y is not searched so Y::f is not considered)
              // Overload resolution picks A::f(int)
 
    AB::x++;  // First, AB is searched. There is no x
              // Then A, B are searched. There is no x
              // Then Y and Z are searched. There is still no x: this is an error
 
    AB::i++;  // AB is searched. There is no i
              // Then A, B are searched. A::i and B::i found by lookup: this is an error
 
    AB::h(16.8); // First, AB is searched. There is no h
                 // Then A, B are searched. There is no h
                 // Then Y and Z are searched
                 // Lookup finds Y::h and Z::h. Overload resolution picks Z::h(double)
}

namespace A { int a; }
 
namespace B { using namespace A; }
 
namespace D { using A::a; }
 
namespace BD
{
    using namespace B;
    using namespace D;
}
 
void g()
{
    BD::a++; // OK: finds the same A::a through B and through D
}