[英]Why is movement constructor invoked twice while std::emplace_back is called only once?
I understand that std::emplace_back
uses placement-new to construct the element in-place at the location provided by the container.我知道
std::emplace_back
使用placement-new 在容器提供的位置就地构造元素。
Why is movement constructor invoked twice while std::emplace_back
is called only once?为什么移动构造函数被调用两次,而
std::emplace_back
只被调用一次?
Here is the related code(check https://godbolt.org/z/-NXzNY ):这是相关代码(检查https://godbolt.org/z/-NXzNY ):
#include <vector>
#include <string>
#include <iostream>
struct President
{
std::string name;
std::string country;
int year;
President(std::string p_name, std::string p_country, int p_year)
: name(std::move(p_name)), country(std::move(p_country)), year(p_year)
{
std::cout << "I am being constructed.\n";
}
President(President&& other)
: name(std::move(other.name)), country(std::move(other.country)), year(other.year)
{
std::cout << "I am being moved.\n";
}
President& operator=(const President& other) = default;
};
int main()
{
std::vector<President> elections;
std::cout << "emplace_back:\n";
elections.emplace_back("Nelson Mandela", "South Africa", 1994);
President pst("Franklin", "the USA", 1936);
std::cout << "=====================" << std::endl;
elections.emplace_back(std::move(pst));
}
You have two objects;你有两个对象; Putting the second to the vector causes a vector resize, memory reallocation, hence the movement requirement.
将第二个放入向量会导致向量调整大小,memory 重新分配,因此需要移动。 In addition to construction-in-place that
emplace_back
does, that's another point of the move construction: To avoid expensive copies when std::vector
resizes.除了
emplace_back
所做的就地构造之外,这是移动构造的另一点:在std::vector
调整大小时避免昂贵的副本。
When adding this line to your code:将此行添加到您的代码时:
elections.reserve(2);
Then you have only one movement.那么你只有一个动作。
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