有效地将 std::unordered_set 的内容移动到 std::vector
在我的代码中,我有一个 std::unordered_set
,我需要将数据移动到 std::vector
中.我在获取数据时使用 std::unordered_set
以确保在转换为 std::vector
之前只存储唯一值.我的问题是如何最有效地将内容移动到 std::vector
?移动数据后我不需要 std::unordered_set
.我目前有以下:
In my code I have a std::unordered_set
and I need to move the data into a std::vector
. I'm using the std::unordered_set
while getting the data to ensure only unique values are stored prior to converting to a std::vector
. My question is how do I move the contents to the std::vector
the most efficiently? I don't need the std::unordered_set
after the data is moved. I currently have the following:
std::copy(set.begin(), set.end(), std::back_inserter(vector));
推荐答案
在C++17之前,你能做的最好的就是:
Before C++17, the best you can do is:
vector.insert(vector.end(), set.begin(), set.end());
set
的元素是 const
,所以你不能离开它们――移动只是复制.
The set
's elements are const
, so you can't move from them - moving is just copying.
在 C++17 之后,我们得到 extract()代码>:
After C++17, we get extract()
:
vector.reserve(set.size());
for (auto it = set.begin(); it != set.end(); ) {
vector.push_back(std::move(set.extract(it++).value()));
}
尽管您的评论是您的数据是 double
s,但这并不重要.
Although given your comment that your data is double
s, this wouldn't matter.
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