C ++函数返回一个右值，但是可以分配一个新值？ [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 ++编译器出现以下错误：

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&)’

它说我不能传递一个rvalue作为一个非const引用的参数，但我困惑的是为什么我

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

传递右值 rtByValue / code>到期望一个左值引用的函数不工作，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lvalue引用–我不会全部引用它，但它基本上说可以从另一个左值引用初始化一个左值引用

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

< >

• 或者可以转换为中间类型


• 或右值的左值引用，但只有当我们初始化的左值引用是const-引用

由于我们需要初始化的参数不是const引用，所以这一切都不适用。

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）为了修改对象，对象的左值是必要的，除了类类型的右值也可以用来修改它的指针情况。 [示例：为对象（9.3）调用的成员函数可以修改对象。 - end example]

(§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 类型，并且（隐式定义的）赋值运算符是成员函数。 （有关详情，请参见此相关问题 。）

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 ，那么分配将不起作用。）

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

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