我有两个向量:
std::vector<int> v1, v2;
// Filling v1
...
现在我需要将 v1
复制到 v2
.有什么理由更喜欢
And now I need to copy v1
to v2
. Is there any reason to prefer
v2 = v1;
到
std::copy (v1.begin(), v1.end(), v2.begin());
(反之亦然)?
一般来说我更喜欢 v2 = v1
:
v2
与 v1
的长度不同,std::copy
将不起作用(它不会调整它的大小),所以它会在最好的情况下保留一些旧元素(v2.size() > v1.size()
并覆盖程序最坏情况下其他地方使用的一些随机数据v1
即将过期(并且您使用 C++11),您可以轻松修改它以移动
内容std::copy
慢,因为实施者可能会在内部使用 std::copy
,如果它能带来性能优势.std::copy
won't work if v2
doesn't have the same length as v1
(it won't resize it, so it will retain some of the old elements best case (v2.size() > v1.size()
and overwrite some random data used elsewhere in the program worst casev1
is about to expire (and you use C++11) you can easily modify it to move
the contentsstd::copy
, since the implementers would probably use std::copy
internally, if it gave a performance benefit.总而言之,std::copy
的表现力较差,可能会做错事,甚至更快.所以真的没有任何理由在这里使用它.
In conclusion, std::copy
is less expressive, might do the wrong thing and isn't even faster. So there isn't really any reason to use it here.
这篇关于复制 std::vector:更喜欢赋值还是 std::copy?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!