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C++的Name lookup之qualified name lookup

編輯：關於C++

1. 簡介

C++中Name lookup的過程就是一個通過name找到對應申明的過程。對於函數來說，name lookup可能會匹配出很多個申明，然後如果是函數再通過Argument-dependent lookup去查找，如果是函數模板那麼通過Template argument deduction去查找，最後一步通過overload resolution去解決。

name lookup分為兩種，qualified name lookup和unqualified name lookup，第一個name的左邊是否還有::符號。

2. qualified name lookup

一個Qualified name出現在::的右邊，它可以包括：類成員（方法，類型，模板等）；命名空間成員（另一個命名空間）；枚舉。

規則如下：

::左邊如果沒有其他元素的話，那麼默認使用頂級命名空間，示例如下:

#include 
int main()
{
  struct std{};
  std::cout << fail
; // Error: unqualified lookup for 'std' finds the struct
  ::std::cout << ok
; // OK: ::std finds the namespace std
}

namelookup的順序是從左到右的，::左邊的元素只會被當做命令空間，類，枚舉，模板,示例如下：

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
}

如果一個qualified name作為一個一個聲明，那麼在它之後的所有相同申明的name的lookup都會和它一樣，在他之前的除外。

class X { };
constexpr int number = 100;
class C {
  class X { };
  static const int number = 50;
  static X arr[number];
};
X C::arr[number], brr[number];   // Error
  // Every name in the declarator C::arr[number] after C::arr
  // is looked up within C::, but the names before C::arr are unaffected,
  // The names in the second declarator (brr[number]) are also unaffected
  // equivalent to:
  // ::X C::arr[C::number], brr[::number]
C::X C::arr[number], brr[number]; // Compiles, size of arr is 50, size of brr is 100

如果::的右邊是一個析構函數，那麼這個析構函數的name lookup將會和::左邊的name lookup保持一致。

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
}

如果：：的左右兩邊都是相同的類名，那麼他只能被當做一個構造函數。但是當使用Elaborated type specifier時後面的函數就會被忽略掉。

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)

qualified name lookup可以用來訪問被嵌套或者派生隱藏的類成員

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)
}

當：：的左邊指定了命名空間，或者：：左邊沒有name（使用頂級命名空間），：：的右邊就使用該命名空間，除了下面這種情況，即name被用作模板參數。

namespace N {
    template struct foo {};
    struct X {};
}
N::foo x; // error: X is looked up as ::X, not as N::X

在命名中間嵌套引用中，Qualified name lookup遞歸執行，先在第一層中lookup，如果沒有就在引用的命名空間中lookup，一直遞歸下去：

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)
}

而且允許存在相同的申明

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
}