Ostream<重载混乱 [英] Ostream << overloading confusion
问题描述
当您重载<< （假定这是在SomeClass中定义为一个朋友），为什么你们都参考ostream并返回ostream？
When you're overloading the << operator for a class (pretend this is defined as a friend in SomeClass), why do you both take a reference to the ostream and return that ostream?
ostream& operator<<(ostream& s, const SomeClass& c) {
//whatever
return s;
}
当ostream已经可以通过引用直接修改时，这看起来对我来说是多余的 - 虽然我确定不是：）
What benefit can returning the ostream be when it was already directly modifiable by reference? This seems redundant to me - though I'm sure it's not :)
推荐答案
这不是多余的， 。使用std :: string :: append这样的函数更容易看到，所以我将从这里开始：
This is not redundant, but useful for chaining calls. It's easier to see with functions like std::string::append, so I'll start with that:
std::string mystring("first");
mystring.append(" second");
mystring.append(" third");
可以重写为：
std::string mystring("first").append(" second").append(" third");
这是可能的，因为.append（）返回对它修改的字符串的引用，添加.append（...）到结束。与你正在做的事情相关的代码正在改变：
This is possible because .append() returns a reference to the string it modified, so we can keep adding .append(...) to the end. The code correlating to what you are doing is changing from this:
std::cout << "first";
std::cout << " second";
std::cout << " third";
因为运算符<返回流，我们也可以链接这些！
into this. Since operator<< returns the stream, we can also chain these!
std::cout << "first" << " second" << " third";
看到相似性和有用性？
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