April 2018
Intermediate to advanced
303 pages
4h 54m
Chinese
Content preview from Effective Modern C++（中文版）
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lambda
表达式
｜
219
Widget&&
&& forward(
Widget
& param)
// T
取值
Widget&&
时
{
// std::forward
的实例化结果
return static_cast<
Widget&&
&&>(param);
//
（在引用折叠发生之前）
}
然后，应用引用折叠规则，右值引用的右值引用结果是单个右值引用，实例化结果为：
Widget&&
forward(Widget& param)
// T
取值
Widget&&
时
{
// std::forward
的实例化结果
return static_cast<
Widget&&
>(param);
//
（在引用折叠发生之后）
}
如果你把这个实例化和在
T
为
Widget
时调用的
std::forward
那个实例化两相比较，
你会发现它们别无二致。那就意味着，实例化
std::forward
时，使用一个右值引用型
别和使用一个非引用型别，会产生相同结果。
这个结果非常不错，因为如果传递给我们的
lambda
式的形参
x
是个右 值，
decltype(x)
产生的是右值引用型别。我们之前已经知道了，传递左值给我们的
lambda
式时，
decltype(x)
会产生传递给
std::forward
的符合惯例的型别，而
现在我们又知道对于右值而言，虽然说
decltype(x)
产生的型别并不符合传递给
std::forward
的型别形参的惯例，但是产生的结果与符合惯例的型别殊途同归。 ...
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