矢量push_back移动实现

问题描述

在我的课本中，矢量push_back move实现的实现是:

In my textbook, the implementation of the vector push_back move implementation is:

void push_back( Object && x )
{
    if( theSize == theCapacity )
        reserve( 2 * theCapacity + 1 );
    objects[ theSize++ ] = std::move( x );
}

我对std :: move的理解是，它基本上是静态的，将项目强制转换为右值引用.那么，为什么在最后一行中，当x已经作为右值引用传入时，他们必须使用std :: move(x)?

My understanding of std::move is that it basically static casts the item as an rvalue reference. So why on the last line did they have to use std::move( x ), when x was passed in already as an rvalue reference?

推荐答案

x 是右值引用，但必须遵循的经验法则是:如果有名称，则为左值 .因此，您必须应用 std :: move 将其类型转换为右值.如果您省略了 std :: move ，则将复制 x 而不是将其移至目标位置.可以在已解释的右值引用中找到更多信息.

x is an rvalue reference, but the rule of thumb you must follow is: if it has a name, it's an lvalue. Therefore you must apply std::move to convert its type to an rvalue. If you left out the std::move then x would be copied instead of moved into its destination. More information can be found in Rvalue References Explained.

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