运算符<<重载ostream
为了这样使用 cout ： std::cout << myObject，为什么我必须传递 ostream 对象？我认为这是一个隐式参数。
ostream &operator<<(ostream &out, const myClass &o) {
out << o.fname << " " << o.lname;
return out;
}
谢谢
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您不会向
ostream添加另一个成员函数，因为这需要重新定义该类。您无法将其添加到myClass中，因为ostream首先出现。您唯一能做的就是向独立函数添加重载，这就是您在示例中所做的。You aren't adding another member function to
ostream, since that would require redefining the class. You can't add it tomyClass, since theostreamgoes first. The only thing you can do is add an overload to an independent function, which is what you're doing in the example.仅当它是类的成员函数时才会成为第一个参数。因此，它会是：
由于
ostream是在您的课程之前很久编写的，因此您会看到将您的课程放在那里的问题。因此，我们必须将运算符实现为独立函数：当运算符实现为类的成员函数时，左侧是
this，参数变为右侧。 （对于二元运算符 - 一元运算符的工作原理类似。）Only if it is a member function of the class that would otherwise be the first argument. Thus, it would be:
Since
ostreamwas written long before your class, you see the problem of getting your class in there. Thus, we must implement the operator as a freestanding function:When operators are implemented as member-functions of classes, the left hand side is
this, and the argument becomes the right hand side. (For binary operators - unary operators work similarly.)因为您正在重载自由函数，而不是成员函数。
Because you are overloading a free function, not a member function.