如何可以分配返回的对象？ [英] How can a returned object be assignable?

问题描述

在Effective C ++中，项目3，Scott Meyers建议为 Rational ：重载 operator * p>

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

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

 （a * b）= c; 
  

或更可能：

  if（a * b = c）
  

现在我很困惑，因为我认为函数的返回值，这里是operator *，是一个右值，因此不可分配。我认为它不可分配，因为如果我有：

  int foo（）; 
 foo（）+ = 3; 
 

无法编译赋值中的无效左值。
为什么不会发生这里？

：我在Scott Meyers的这个项目上看到过很多其他的线程，但是没有解决这个问题。 。

解决方案

关于类类型， a = b 只是 a.operator =（b）的简写，其中 operator = 功能。

请注意，在C ++ 11中，你可以通过 operator = lvalue-only：

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

& 告诉编译器函数不能在右值上调用。

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);

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;

or, more probably:

     if (a*b = c)

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;

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

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

解决方案

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.

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

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

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

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