为什么'std :: endl'在语句'std :: cout<< std :: endl;“，给定参数相关的查找？

问题描述：

我在查看与参数相关的查询的维基百科词条，以及（Jan 04，2014）给出了以下示例：

I was looking at the Wikipedia entry on argument-dependent lookup, and (on Jan 04, 2014) the following example was given:

#include<iostream>

int main() 
{
  std::cout << "Hello World, where did operator<<() come from?" << std::endl;
}

...具有以下注释：

... with the following comment:

注意，std :: endl是一个函数，但它需要完全限定，
，因为它被用作运算符<< （std :: endl是一个函数
指针，而不是函数调用）。

Note that std::endl is a function but it needs full qualification, since it is used as an argument to operator<< (std::endl is a function pointer, not a function call).

我的想法是不正确（或简单不清楚）。我正在考虑更改注释，而是

My thought is that the comment is incorrect (or simply unclear). I am considering changing the comment to say, instead

请注意，std :: endl需要完全限定，
因为ADL不应用于函数调用的参数; it
仅适用于函数名称本身。

Note that std::endl needs full qualification, because ADL does not apply to the arguments of a function call; it only applies to the function name itself.

我是否纠正维基百科的评论不正确？我的建议更改是否正确？（

Am I correct that the Wikipedia comment is incorrect? Is my proposed change correct? (I.e., is my understanding of ADL correct in this example?)

答

维基百科说的没什么问题。

There's nothing wrong about what Wikipedia says.

std::cout << "Hello World, where did operator<<() come from?" << std::endl

相当于以下内容（假设 operator 实现为免费函数）

is equivalent to the following (assuming operator<< is implemented as a free function)

operator<<(
    operator<<(std::cout, "Hello World, where did operator<<() come from?"),
    std::endl)

这显然需要 cout 和 endl 的命名空间限定这是参数相关的查找（函数），而不是参数查找。

参数确定要调用的函数，而不是

which clearly requires namespace qualification for both cout and endl because this is argument-dependent lookup (of the function), not "argument lookup".
The arguments determine the function to be called, not the way around.

为什么'std :: endl'在语句'std :: cout<< std :: endl;“，给定参数相关的查找？

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