为什么GCC允许调用这个函数而不首先使用它的命名空间？ [英] Why GCC allows calling this function without using its namespace first?

问题描述

可能重复：

Possible Duplicate:
Why does C++ parameter scope affect function lookup within a namespace?

今天我经历了这种奇怪的行为。我可以使用命名空间Strange 调用strangeFn，但不允许调用strangeFn2为什么？

Today I experienced this weird behavior. I can call strangeFn without using namespace Strange first, but does not allow calling strangeFn2 Why?

namespace Strange
{
    struct X
    {
    };
    void strangeFn(X&) {}
    void strangeFn2(int) {}
}

int main()
{
    Strange::X x;
    strangeFn(x);    // GCC allows calling this function.
    strangeFn2(0);   // Error: strangeFn2 is not declared in this scope.
    return 0;
}

C ++编译器如何解析符号的范围？

How does C++ compilers resolve the scope of the symbols?

推荐答案

这称为 Argument Dependent Lookup（或Koenig Lookup）

This is called Argument Dependent Lookup (or Koenig Lookup)

基本上，如果一个符号无法解析，

Basically, if a symbol couldn't be resolved, the compiler will look into the namespace(s) of the argument(s).

第二个函数调用失败，因为 strangeFn2 isn' t在当前命名空间中可见，它在参数类型的命名空间中未定义（ int ）

The second function call fails, because strangeFn2 isn't visible in the current namespace, neither is it defined in the namespace of it's parameter type (int)

可以看到这对于操作符函数是否有效：

You can see how this works well with operator functions:

 std::complex<double> c, d;
 c += d; // wouldn't really work without ADL

或无所不在的iostream运算符：

or the ubiquitous iostream operators:

 std::string s("hello world");
 std::cout << s << std::endl; // Hello world would not compile without ADL...

有趣的是，这就是hello world看起来像没有ADL （且没有使用关键字...）：

For fun, this is what hello world would look like without ADL (and without using keyword...):

 std::string s("hello world");
 std::operator<<(std::cout, s).operator<<(std::endl); // ugly!

在功能模板存在的情况下，有ADL和重载分辨率的阴影角落情况，现在让他们离开答案的范围。

There are shadowy corner cases with ADL and overload resolution in the presence of function templates, but I'll leave them outside the scope of the answer for now.

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