返回的对象如何可分配?-C/C++问题

How can a returned object be assignable?(返回的对象如何可分配?)

本文介绍了返回的对象如何可分配?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

在 Effective C++ 第 3 条中，Scott Meyers 建议为名为 Rational 的类重载 operator*:

In Effective C++, Item 3, Scott Meyers suggests overloading operator* for a class named Rational:

    class Rational { ... };
    const Rational operator*(const Rational& lhs, const Rational& rhs);

返回值是const-qualified的原因解释如下:如果不是const，程序员可以编写如下代码:

The reason for the return value being const-qualified is explained in the following line: if it were not const, programmers could write code such as:

    (a * b) = c;

或者，更有可能:

     if (a*b = c)

很公平.现在我很困惑，因为我认为函数的返回值(此处为 operator*)是一个右值，因此不可赋值.我认为它不可分配，因为如果我有:

Fair enough. Now I’m confused as I thought that the return value of a function, here operator*, was a rvalue, therefore not assignable. I take it not being assignable because if I had:

    int foo();
    foo() += 3;

使用 invalid lvalue in assignment 将无法编译.为什么这里不会发生?有人可以对此有所了解吗?

that would fail to compile with invalid lvalue in assignment. Why doesn’t that happen here? Can someone shed some light on this?

编辑:我在 Scott Meyers 的那个项目上看到了许多其他线程，但没有一个解决我在这里暴露的右值问题.

EDIT: I have seen many other threads on that very Item of Scott Meyers, but none tackled the rvalue problem I exposed here.

推荐答案

重点是对于类类型，a = b 只是 a.operator=(b) 的简写，其中 operator= 是成员函数.并且可以在右值上调用成员函数.

The point is that for class types, a = b is just a shorthand to a.operator=(b), where operator= is a member function. And member functions can be called on rvalues.

请注意，在 C++11 中，您可以通过使 operator= 仅左值来禁止它:

Note that in C++11 you can inhibit that by making operator= lvalue-only:

class Rational
{
public:
  Rational& operator=(Rational const& other) &;
  // ...
};

& 告诉编译器这个函数不能在右值上调用.

The & tells the compiler that this function may not be called on rvalues.

这篇关于返回的对象如何可分配?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！