>>和<<操作符重载 [英] >> and << operator overloading

问题描述

我刚刚为我的编程课做了一个测验，并得到了这个问题：

I just did a quiz for my programming class and got this question wrong:

函数的返回类型为
重载操作符< 必须是对ostream对象的
引用。

The return type of the function to overload the operator << must be a reference to an ostream object.

这对我来说似乎不对。当然C ++有点比这更开放。但我想我会问这里。这是正确的（或错误）？我的C ++知识开始真正褪色，当涉及到操作符重载..

This does not seem right at all to me. Surely C++ is a bit more open ended than this. But I thought I'd ask here anyway. How is this right (or wrong)? My C++ knowledge begins to really fade when it comes to operator overloading..

推荐答案

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.

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