重载的<< [英] Return value of overloaded <<

问题描述

#include <iostream>

using namespace std;

struct info {
    info(int x, int y) : x(x), y(y) {}
    int x;
    int y;
};

ostream& operator<<(ostream& out, const info &myinfo){
    out << myinfo.x << "  " << myinfo.y;
    return cout;
}

int main() {
    info a(1,2);
    info b(3,4);
    cout << a << " " << b << endl;
}

上述程序的输出似乎很好，即使有运算符<< 。

The output of the above program seems fine even with the incorrect overload of operator <<.

任何人都可以告诉我这个重载问题的影响是什么？我知道重载函数应该返回 out 而不是 cout ，但是上面的版本如何行为？

Can anyone tell me what is the effect of this overloading problem? I know the overloading function should return out instead of cout, but how does the above version behave?

推荐答案

在这种情况下，由于你传递 std :: cout operator<< ，行为上没有差别。一般来说，你会导致<< b<< std :: endl 以发送到 std：cout ，而您的 a

In this case, since you are passing in std::cout to the overloaded operator<<, there is no difference in behavior. Generally, though, you would cause the " " << b << std::endl to get sent to std:cout, while your a would go to whatever you passed in.

例如：

info a(1,2);
info b(3,4);
std::ostringstream ss;
ss << a << " " << b << std::endl;

会导致 a 转到 $ 。

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