C++ 函数返回一个右值，但可以为其分配一个新值? [英] C++ function returns a rvalue, but that can be assigned a new value?

问题描述

代码如下:

 #include <iostream>
 using namespace std;

 class A {

 };

 A rtByValue() {
return A();
 }

 void passByRef(A &aRef) {
    // do nothing
 }

 int main() {
    A aa;
    rtByValue() = aa;            // compile without errors
    passByRef(rtByValue());      // compile with error 

    return 0;
 }

g++ 编译器给出以下错误:

The g++ compiler gives the following error:

d.cpp: In function ‘int main()’:
d.cpp:19:23: error: invalid initialization of non-const reference of type ‘A&’ from an rvalue of type ‘A’
d.cpp:12:6: error: in passing argument 1 of ‘void passByRef(A&)’

它说我不能将右值作为非常量引用的参数传递，但我感到困惑的是为什么我可以分配给这个右值，正如代码所示.

It says that I can't pass an rvalue as an argument of a non-const reference, but what I'm confused about is why I can assign to this rvalue, just as the code shows.

推荐答案

将右值 rtByValue() 传递给需要左值引用的函数不起作用，因为这将需要左值引用参数从右值初始化.§8.5.3/5 描述了如何初始化左值引用——我不会完整引用它，但它基本上是说可以初始化左值引用

Passing the rvalue rtByValue() to a function that expects an lvalue reference doesn't work because this would require the lvalue reference argument to be initialized from an rvalue. §8.5.3/5 describes how lvalue references can be initialized – I won't quote it in full, but it basically says that an lvalue reference can be initialized

• 来自另一个左值引用
• 或者可以转换为中间类型的左值引用的东西
• 或来自右值，但前提是我们初始化的左值引用是常量引用

由于我们需要初始化的参数不是常量引用，所以这些都不适用.

Since the argument we need to initialize is not a const-reference, none of this applies.

另一方面，

rtByValue() = aa; 

即分配给临时对象是可能的，因为:

i.e., assigning to a temporary object, is possible because of:

(§3.10/5) 对象的左值对于修改对象是必要的，除了在某些情况下也可以使用类类型的右值来修改其所指对象.[ 示例:为对象调用的成员函数(9.3)可以修改该对象.— 结束示例 ]

(§3.10/5) An lvalue for an object is necessary in order to modify the object except that an rvalue of class type can also be used to modify its referent under certain circumstances. [ Example: a member function called for an object (9.3) can modify the object. — end example ]

所以这只是因为 A 是类类型的，并且(隐式定义的)赋值运算符是一个成员函数.(请参阅此相关问题了解更多详情.)

So this works only because A is of class-type, and the (implicitly defined) assignment operator is a member function. (See this related question for further details.)

(因此，如果 rtByValue() 返回，例如，int，则赋值将不起作用.)

(So, if rtByValue() were to return, for example, an int, then the assignment wouldn't work.)

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