如何<<操作员工作正常？ [英] How does the << Operator exactly work?

问题描述

我在理解<<

I have some problems in understanding the << operator.

如果我有：

#include <iostream>
using namespace std;
//...

int t = 5;
cout << "test is: " << t << endl;

现在函数运算符<<叫做。

Now the function operator<< is called.

ostream& operator<<(ostream& out, string* s)
{
    return out << s << endl;
}


ostream& operator<<(ostream& out, int* value)
{
    return out << value << endl;
}

字符串指针指向具有值 引用（ cout ？）并且是ostream&

the string-pointer points to the address with value test is: but to what does the element out refer (to cout?)? and is the function body of ostream& correct in that way?

推荐答案

首先，我们修复你的代码：运算符应该使用 const 引用或值，而不是指针：

First, let's fix your code: the operators should be taking const references or values instead of pointers:

ostream& operator<<(ostream& out, const string& s) // const reference
ostream& operator<<(ostream& out, int i)           // value

是正确的， out 参数接收对 cout 的引用，或者 ostream& ; << 左侧的表达式返回。尽管 - < 左侧的表达式不一定是 cout << 运算符 * 用于链接和流操纵器。在所有情况下，这些表达式返回对 ostream 的引用，以便chain可以继续。

Now to your question: you are correct, the out parameter receives the reference to the cout, or whatever is the ostream& returned from the expression on the left side of <<. The expression on the left of << is not necessarily cout, though - other common cases are results of other << operators^* for chaining, and stream manipulators. In all cases these expressions return a reference to ostream so that the "chain" could continue.

* 运算符<< 返回 ostream& 的原因是可以链接输出。在绝大多数情况下，您会返回与接收作为第一个参数相同的 ostream& ，但是对标准C ++库要求您做的部分没有限制

^* The reason the operator<< return an ostream& is so that you could chain the output. In overwhelming number of cases you wold return the same ostream& that you receive as the first parameter, although there is no limitation on the part of the standard C++ library requiring you to do that.

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