Question

依赖于参数的查找说：

对于类类型的参数（包括union），该集合包含... a）班级本身b）......

那为什么printX找不到X？

 #include<iostream>
using namespace std;

class A {
public:
    static const int X = 1;
};

class B {
public:
    static void printX(A a)
    {
        cout << "X is " << X << endl;
    }
};

int main(int argc, char** argv)
{
    A a;
    B::printX(a);
    return 0;
}

Answer 1

这不是ADL应该做的事情。 ADL适用于非限定函数调用，该函数首先在名称空间中搜索，其中定义了参数的类型

实施例，

namespace N
{
      struct A {};

      void f(A const &) {}
}

N::A a;

N::f(a); //qualified function call. ADL is not needed.
f(a);    //unqualified function call. ADL works here

在上面的示例中f(a)从名称空间f()调用N。由于f(a)是非限定函数调用，因此ADL使编译器能够首先搜索命名空间f中的函数N作为参数的类型{{1} }在a中定义。

希望有所帮助。

Answer 2

我想扩大@Nawaz的答案，它会对的含义有所启发。“该集合包含...... a）类本身”。

示例程序：

#include <iostream>

void bar(int ) { std::cout << "Came to ::bar(int)\n"; }

namespace foo
{
   struct A2;
   struct A1
   {
      friend void bar(A1 const&) { std::cout << "Came to foo::bar(A1 const&)\n"; }
      friend void bar(A2 const&) { std::cout << "Came to foo::bar(A2 const&)\n"; }
   };

   struct A2
   {
   };
}

struct B
{
   friend void bar(B) { std::cout << "Came to ::bar(B)\n"; }
};

int main()
{
   foo::A1 a1;
   foo::A2 a2;
   B b;
   int i = 0;

   bar(i);   // This resolves to the only bar() in the global namespace.

   bar(a1);  // This resolves to the appropriate overloaded friend function
             // defined in foo::A1 because the scopes for looking up the 
             // function includes class foo::A1

   bar(a2);  // Error. The search for bar include foo::A2 but not foo::A1.        

   bar(b);   // This resolves to the friend function defined in B because
             // the scopes for looking up the function includes class B
}

为什么Koening查找不能在这里工作？

2 个答案: