将 std::endl 传递给 std::operator <<

Question

在这个Stack Overflow answer 它说 std::cout << "Hello World!" << std::endl; 与

相同

std::operator<<(std::operator<<(std::cout, "Hello World!"), std::endl);

但是当我编译上面这行代码的时候，编译不通过！然后在尝试其他方法后我发现它不编译的原因是因为 std::endl，如果我用 "\n" 替换 std::endl 然后它工作。但是为什么不能把std::endl传给std::operator<<呢？

或者更简单地说，std::cout<<std::endl; 和 std::operator<<(std::cout, std::endl); 不一样吗？

编辑

用icpc test.cpp编译时，报错信息是 error: no instance of overloaded function "std::operator<<" matches the argument list argument types are: (std::ostream, <unknown-type>) std::operator<<(std::cout, std::endl);

和 g++ test.cpp 给出了更长的错误消息。

Answer 1

我不知道这个话题，但我认为这 2 个问题和答案与您的问题有些相关，可能会帮助您找到解决方案

operator << must take exactly one argument

Does std::cout have a return value?

Answer 2

因为那里的答案有点不对。 std::endl is a manipulator function, there is no overload for them in definitions of standalone operator<< of ostream. It is a member function of basic_ostream.

换句话说，目前的调用是错误的。它应该是以下之一：

#include <iostream>
int main() {
    std::endl(std::operator<<(std::cout, "Hello World!"));
    std::operator<<(std::cout, "Hello World!").operator<<(std::endl);

    //of course if you pass new line as a supported type it works
    std::operator<<(std::operator<<(std::cout, "Hello World!"), '\n');
    std::operator<<(std::operator<<(std::cout, "Hello World!"), "\n");
    std::operator<<(std::operator<<(std::cout, "Hello World!"), string("\n"));
    return 0;
}

Live examples.

嗯，确实有人说流库没有标准中最漂亮的设计。

将 std::endl 传递给 std::operator <<

pass std::endl to std::operator <<

c++

std