问题描述
当打印一个对象时，一个朋友操作符<<用来。我们可以使用成员函数for operator<< ？
When you are going to print an object, a friend operator<< is used. Can we use member function for operator<< ?
class A {
public:
void operator<<(ostream& i) { i<<"Member function";}
friend ostream& operator<<(ostream& i, A& a) { i<<"operator<<"; return i;}
};
int main () {
A a;
A b;
A c;
cout<<a<<b<<c<<endl;
a<<cout;
return 0;
}
一点是，friend函数使我们能够像这样使用
One point is that friend function enable us to use it like this
cout<<a<<b<<c
其他原因？
推荐答案
不是二元运算符的成员函数，右侧的成员函数作为另一个参数传递，左侧始终为 * this 。
You have to use a free function and not a member function as for binary operators the left hand side is always *this for member functions with the right hand side being passed as the other parameter.
对于输出流操作符，左手边总是流对象，所以如果你流到一个标准类，而不是自己编写流你必须提供一个自由功能，而不是你的成员类。
For output stream operators the left hand side is always the stream object so if you are streaming to a standard class and not writing the stream yourself you have to provide a free function and not a member of your class.
虽然可以提供一个后向流操作符作为成员函数，并且流如下：
Although it would be possible to provide a backwards stream operator as a member function and stream out like this:
myObject >> std::cout;
不仅会违反一个非常强的库惯例，正如你所指出的，链接输出操作不会由于从左到右分组>> ，工作。
not only would you violate a very strong library convention, as you point out, chaining output operations would not work due to the left-to-right grouping of >>.
/ strong>正如其他人所说，虽然你必须使它成为一个自由函数，只需要一个朋友如果流函数不能实现类的'公共接口。
As others have noted, while you have to make it a free function it only needs to be a friend if the streaming function cannot be implemented in terms of the class' public interface.
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