C ++不要求返回类型是对 ostream 对象的引用。但是，如果您尝试执行以下操作：

It is not required by C++ that the return type be a reference to an ostream object. However, if you are trying to do something like:

cout << instance_of_custom_type << 3 << "hi" << endl;

然后您将需要：

ostream &operator << (ostream &os, custom_type &t);

然而，如果你正在做类似于编写大型整数类型，它可能是这样的：

However, if you were doing something like writing a large integer type, and wanted to support bit shifting, it might be something like:

BigInt operator << (const BigInt &i, unsigned int shift);

要进一步扩展， << 运算符用于位移。 1<< 8 为256。 C ++为此添加了一个（稍微混乱的）第二次使用，并在 ostream 上重载它以表示输出到流。你可以在一个重载的操作符里面做任何你喜欢的事情 - 它的作用就像一个函数，然而，操作符有一个人类的期望与他们附加：程序员期望，在C + +，< 是位移位或流输出。

To expand this a bit further, the original use of the << operator is for bit shifting. 1 << 8 is 256, for example. C++ added a (slightly confusing) second use for this, and overloaded it on ostream to mean "output" to the stream. You can do whatever you like within an overloaded operator - it works just like a function, however, operators have a human expectation attached with them: programmers expect, in C++, that << is bit shifting or stream output.