如果通过 const 引用延长了它的生命周期，那么这个右值在哪里？

Question

我已经通读了这个问题：

Where is rvalue stored in C

并根据答案：

int main()
{
    int a = 7; // 7 rvalue is stored in the program binary and assigned to "a"
    int b = 7 * rand(); // 7 is pulled from the program binary, multiplied by result of rand and assigned to "b"

    // But...
    const int& c = 3; // 3 is an rvalue?
    const_cast<int&>(c) = 1; // rvalues don't have storage space? But I'm assigning to it.
    std::cout << c; // Prints 1
}

我想了解，通常右值临时（等号右侧）在内存中没有存储 space 并且其生命周期在行尾结束。如果我们将一个 const 引用绑定到它，那么它会将右值的生命周期延长到它所绑定的引用的生命周期。在分配给数字文字“3”的情况下，这是否意味着它确实获得了一些存储空间 space 而不是右值？还是右值？

另外我想知道我所做的是否是未定义的行为，无论 const_cast 是不可取的还是其他什么。

Answer 1

normally a rvalue temporary (right side of the equals sign) doesn't have storage space in memory

正常情况下是有存储的space，你应该是理解错了一些概念吧

In the case to assigning to "3", a number literal, does this mean it does get some storage space and isn't an rvalue? Or is an rvalue?

两者兼而有之：它是一个右值并且它有一个存储space。栈上有a space分配给store和整数，值为3，然后c绑定到这个变量。

关于您的代码：

const int& c = 3; // 3 is an rvalue?

是的，是右值。

const_cast<int&>(c) = 1; // rvalues don't have storage space? But I'm assigning to it.
std::cout << c; // Prints 1

来自const_cast：

const_cast makes it possible to form a reference or pointer to non-const type that is actually referring to a const object or a reference or pointer to non-volatile type that is actually referring to a volatile object. Modifying a const object through a non-const access path and referring to a volatile object through a non-volatile glvalue results in undefined behavior.

如果通过 const 引用延长了它的生命周期，那么这个右值在哪里？

Where is this rvalue residing if its lifetime is prolonged by const reference?

c++

constants

rvalue-reference