免费运营商与会员运营商 [英] Free Operators versus Member Operators

问题描述

class Currency
{
public:
    explicit Currency(unsigned int value);
    // method form of operator+=
    Currency &operator +=(const Currency &other); // understood!
    ...
};

以下代码显示了使用免费版本的运算符的等效API:

The following code shows an equivalent API using a free function version of the operator:

class Currency
{
public:
    explicit Currency(unsigned int value);
    ...
};

// free function form of operator+=
Currency &operator +=(Currency &lhs, const Currency &rhs); // ???

Question1 为什么自由函数应该返回Currency&而不是Currency? 这是一个好习惯吗?

Question1> Why should the free function return Currency& instead of Currency? Is this a good practice?

Question2 在实现中，应该使用哪个变量返回lhs或rhs?

Question2> In the implementation, which variable should be used to return, lhs or rhs?

推荐答案

operator+=的标准行为是将lhs增加rhs并返回对lhs的引用.

The standard behavior of operator+= is to increment the lhs by the rhs and return a reference to the lhs.

在成员函数中，lhs是调用对象，因此，它应该返回对自身的引用.您似乎期望自由函数的行为不同于成员函数.为什么?

In the member function, the lhs is the calling object, and accordingly, it should return a reference to itself. You seem to expect the free function to behave differently than the member function. Why?

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