"x值具有身份"是什么意思? [英] What does it mean "xvalue has identity"?

问题描述

C ++ 11引入了新的值类别，其中之一是xvalue.

C++11 introduced new value categories, one of them is xvalue.

Stroustrup对它进行了解释，类似于(im类别):它是一个值，具有身份，但可以从"移出.

It is explained by Stroustrup as something like (im category): "it is a value, which has identity, but can be moved from".

另一个来源， cppreference 解释:

Another source, cppreference explains:

glvalue是一个表达式，其求值确定对象，位字段或函数的身份；

a glvalue is an expression whose evaluation determines the identity of an object, bit-field, or function;

xvalue是glvalue，所以这对xvalue也是正确的.

And xvalue is a glvalue, so this is statement is true for xvalue too.

现在，我认为如果xvalue具有身份，那么我可以检查两个xvalue是否引用同一对象，因此我使用了xvalue的地址.事实证明，这是不允许的:

Now, I thought that if an xvalue has identity, then I can check if two xvalues refer to the same object, so I take the address of an xvalue. As it turned out, it is not allowed:

int main() {
    int a;
    int *b = &std::move(a); // NOT ALLOWED
}

xvalue具有身份是什么意思?

What does it mean that xvalue has identity?

推荐答案

xvalue确实具有标识，但是在语言中有一个单独的规则，即一元&表达式需要一个lvalue操作数.来自[expr.unary.op]:

The xvalue does have an identity, but there's a separate rule in the language that a unary &-expression requires an lvalue operand. From [expr.unary.op]:

一元&运算符的结果是指向其操作数的指针.操作数应为左值[...]

The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue [...]

通过将xvalue绑定到引用，可以在执行rvalue-to-lvalue转换后查看xvalue的身份:

You can look at the identity of an xvalue after performing rvalue-to-lvalue conversion by binding the xvalue to a reference:

int &&r = std::move(a);
int *p = &r;  // OK

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