C ++向量:: clear [英] C++ vector::clear

问题描述

vector <weight> decoy;

void clear_decoy() {  

    decoy.clear();   

    vector<weight> (decoy).swap(decoy);  
}

在上述方法 clear_decoy c $ c>， vector （decoy）.swap（decoy）; 是什么意思？

In the above method clear_decoy(), what does vector<weight> (decoy).swap(decoy); mean please?

方法是否清除 decoy ？感谢！

推荐答案

它创建一个新的向量 Weight 将会是空的），并与 decoy 交换它。

It creates a new vector of Weight objects (which will be empty) and swaps it with decoy.

这样做的原因是默认情况下， std :: vector< t> :: clear 通常不会实际减少向量使用的存储，它只是破坏了那里包含的所有对象。这样，向量就有空间来存储更多对象，而不会在将来重新分配。

The reason for this is that by default, std::vector<t>::clear often doesn't actually reduce the storage used by a vector, it merely destroys all the objects contained there. This way, the vector has room to store more objects without reallocation in the future.

然而，有时候，你想修剪向量中的容量。使用新创建的向量（它生存到它的行尾，并因此被销毁）释放所有由向量分配的内存。

Sometimes, however, you want to trim the capacity in the vector. Swapping with a newly created vector (which lives until the end of it's line, and is therefore destroyed there) frees all the memory allocated by the vector.

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