有必要有一个临时或字面值有一个右值吗？ [英] Is it necessary to have a temporary or a literal to have an rvalue?

问题描述

此问题询问是否所有临时值都是有价值的。

This question asks if all temporaries are rvalue.

答案是否定的，因为如果我们考虑这个表达式：

The answer is no, because if we consider this expression:

const int &ri = 2 + 3;

那么，非常相同的临时（2 + 3），这里是一个右值，可以使用
作为后续表达式中的左值：

then, the very same temporary (2 + 3), which is an rvalue here, can be used as an lvalue in a subsequent expression:

const int *pi = &ri;

所以这个临时值不是（只有）右值。

so this temporary is not (only) an rvalue.

逻辑语句 temporary ==>

The logic statement temporary ==> rvalue is then false.

但是，我们无法写

const int &ri = &(2 + 3); // illegal, 2 + 3 -> temporary -> rvalue

或

int *i = &4; // illegal, 4 is an rvalue (literal)

或

int foo();
int *i = &foo(); // illegal, foo() -> temporary -> rvalue

因此我的问题是，我们可以在某个表达式中生成一个右值，而不是
临时还是文字？是 rvalue ==> （临时或字面） true？

Thus my question is, can we generate an rvalue in a certain expression without having a temporary or a literal? Is rvalue ==> (temporary or literal) true?

推荐答案

强> r值。有一个特殊的规则允许const引用和r值引用绑定到r值，并且这将临时对象的生命周期延长到引用的生命周期（参见12.2（5）），但这不会使临时 - 对象表达式的任何更少的r值。

Expressions that yield temporary objects are r-values. There's a special rule which allows const-references and r-value references to bind to r-values, and this extends the lifetime of the temporary object to that of the reference (see 12.2(5)), but that does not make the temporary-object expression any less of an r-value.

但是，一旦绑定到引用，引用变量本身就有一个名称，因此引用表达式是l -l值。

However, once bound to a reference, the reference variable itself has a name, and thus the reference expression is an l-lvalue.

不要混淆表达式，变量和对象。

Don't confuse expressions, variables and objects.

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