将右值分配给右值 [英] Assigning rvalue to rvalue

问题描述

请问为什么下面的代码可以编译不出错?

struct Test {整数;};int main() {(测试){.num = 10} = (测试){.num = 20};返回0；}

(Test){.num = 10} 和 (Test){.num = 20} 是右值，这是不可能将一个右值分配给另一个右值.cpp 如何认为该语句有效?

PS:代码使用C++11标准编译

解决方案

这就是默认情况下在 C++ 中指定特殊成员函数的方式.例如Test 有一个合成的特殊成员函数:

测试&运算符=(测试常量&)=默认值；//如果你写出来

可以在左值和右值上调用.

如果你想禁止对右值赋值，你必须自定义类的行为:

struct 测试 {整数;测试&operator=(Test const &) &= 默认值；//左值:好的测试&operator=(Test const &) &&= 删除；//右值:错误};

这是 C++20 中的 演示，就像您示例中的代码一样，但相同原则适用于 c++11.

---

请注意，对于内置类型，行为更明智，因此您不能这样做:

42 = 5;3 + 3 = 6;

而且没有办法让它编译.<​​/p>

Can I ask why the code below can be compiled without any error?

struct Test {
    int num;
};

int main() {
    (Test){.num = 10} = (Test){.num = 20};
    return 0;
}

(Test){.num = 10} and (Test){.num = 20} are rvalue, this is impossible that a rvalue can be assigned to another rvalue. How cpp considers the statement is valid?

PS: the code is compiled using C++11 standard

解决方案

This is just how the the special member functions are specified in c++ by default. e.g. Test has a synthesized special member function:

Test& operator=(Test const &) = default; // if you write it out

which can be called on both lvalues and rvalues.

If you want to disallow assignments to rvalues, you have to customize the behavior of the class:

struct Test {
    int num;
    Test& operator=(Test const &) & = default;  // lvalues: ok
    Test& operator=(Test const &) && = delete;  // rvalues: error
};

Here's a demo in c++20, like the code in your example, but the same principle applies in c++11.

---

Note that for in-built types, the behavior is more sensible, so you can't do this:

42 = 5;
3 + 3 = 6; 

and there's no way to let that compile.

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