将右值引用分配给左值引用 [英] assigning Rvalue reference to Lvalue reference
本文介绍了将右值引用分配给左值引用的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
int&& rv = 10;
int& lv = rv; //no error
这怎么可能?
这与引用折叠规则有关吗?
Is this related to "reference collapsing rule"?
推荐答案
int&& rv = 10;
int& lv = rv; //no error
首先,以命名的对象为从不一个右值。其次,由于 rv
是 named 对象,因此即使绑定到rvalue,它也不是右值。由于 rv
是左值,因此可以毫无问题地绑定到左值。
First of all, a named object is never an rvalue. Second, since rv
is named object, it is not a rvalue, even though it binds to rvalue. Since rv
is lvalue, it can bind to lvalue without any problem.
请注意,rvalue-ness是表达式的属性,而不是变量。在上面的示例中,从 10
中创建了一个右值,并将其绑定到 rv
,正如我所说的,它是左值
Note that rvalue-ness is a property of an expression, not a variable. In the above example, an rvalue is created out of 10
and binds to rv
, which as I said, is lvalue.
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