为什么重载ostream的运算符<<需要参考“&"吗? [英] Why does overloading ostream's operator<< need a reference "&"?

问题描述

我一直在学习C ++.

I've been learning C++.

在此页面上，我了解到重载<<"ostream的运算符可以通过这种方式完成.

From this page, I understood that overloading "<<" operator of ostream can be done in this way.

ostream& operator<<(ostream& out, Objects& obj) {
    //
    return out;
} 
//Implementation

和

friend ostream& operator<<(ostream& out, Object& obj);
//In the corresponding header file

我的问题是...为什么此功能需要&"在 ostream 和 Object 的末尾?

My question is... why does this function need "&" at the end of ostream and Object?

至少我知道&"用来...

At least I know that "&" is used to...

1. 获取值的地址
2. 声明对类型的引用

但是，我认为它们均不适用于上述重载.我花了很多时间在谷歌上阅读和阅读教科书，但找不到答案.

However, I think neither of them applies to the overloading described above. I've spent a lot of time in googling and reading a text book, but I can't find the answer.

任何建议将不胜感激.

推荐答案

为什么此功能需要&"在ostream和Object的末尾?

why does this function need "&" at the end of ostream and Object?

因为您要通过引用传递它们.
您为什么要通过引用传递它们.为了防止复制.

Because you are passing them by reference.
Why are you passing them by reference. To prevent a copy.

ostream& operator<<(ostream& out, Objects const& obj)
                             //           ^^^^^       note the const
                             //                       we don't need to modify
                             //                       the obj while printing.

obj 可以被复制(潜在地).但是，如果复制成本很高，该怎么办.因此最好通过引用将其传递，以防止不必要的复制.

The obj can be copied (potentially). But what if it is expensive to copy. So best to pass it by reference to prevent an unnecessary copy.

out 的类型为 std :: ostream .不能复制(复制构造函数已禁用).因此，您需要通过引用.

The out is of type std::ostream. This can not be copied (the copy constructor is disabled). So you need to pass by reference.

我通常在类声明中直接声明流运算符:

I normally declare the stream operator direcctly in the class declaration:

class X
{
    std::string    name;
    int            age;

    void swap(X& other) noexcept
    {
        std::swap(name, other.name);
        std::swap(age,  other.age);
    }
    friend std::ostream& operator<<(std::ostream& str, X const& data)
    {
        return str << data.name << "\n" << age << "\n";
    }
    friend std::istream& operator>>(std::istream& str, X& data)
    {
        X alt;
        // Read into a temporary incase the read fails.
        // This makes sure the original is unchanged on a fail
        if (std::getline(str, alt.name) && str >> alt.age)
        {
            // The read worked.
            // Get rid of the trailing new line.
            // Then swap the alt into the real object.
            std::string ignore;
            std::getline(str, ignore);
            data.swap(alt);
        }
        return str;
    }
};

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