带有复制构造函数的Emplace和Try_emplace
我对EMPLACE和TRY_EMPLECE有一个问题，因为他们在移动对象时始终使用复制构造函数
#include <iostream>
#include <unordered_map>
#include <map>
using namespace std;
class Too {
public:
Too(int x, int y):x_(x), y_(y) {
cout << "init " << x_ << endl;
}
Too(const Too& too):x_(too.x_+1), y_(too.y_+1) {
cout << "Copy happen: x = " << x_ << endl;
}
~Too() {
cout << "delete too " << x_ << endl;
}
private:
int x_, y_;
};
std::map<int, Too> v;
int main()
{
v.emplace(100, Too{ 100,23 });
v.try_emplace(12, 12, 13);
Too t = Too(10, 11);
v.try_emplace(11, std::move(t));
}
。 初始化100 复制发生：x = 101 也删除100 初始化12 初始化10 复制发生：x = 11 也删除10 也删除101 也删除12 如您所见，删除11
，只有v.try_emplace（12，12，13）不使用复制构造函数。 V.Emplace（100，Too {100,23}）和v.try_emplace（11，std :: move（t））调用复制构造函数。
那么，即使我使用std :: Move（t），该怎么办？
任何建议都将不胜感激。
谢谢，
I have an issue with emplace and try_emplace as they always use the copy constructors when moving an object in.
#include <iostream>
#include <unordered_map>
#include <map>
using namespace std;
class Too {
public:
Too(int x, int y):x_(x), y_(y) {
cout << "init " << x_ << endl;
}
Too(const Too& too):x_(too.x_+1), y_(too.y_+1) {
cout << "Copy happen: x = " << x_ << endl;
}
~Too() {
cout << "delete too " << x_ << endl;
}
private:
int x_, y_;
};
std::map<int, Too> v;
int main()
{
v.emplace(100, Too{ 100,23 });
v.try_emplace(12, 12, 13);
Too t = Too(10, 11);
v.try_emplace(11, std::move(t));
}
output
init 100
Copy happen: x = 101
delete too 100
init 12
init 10
Copy happen: x = 11
delete too 10
delete too 101
delete too 12
delete too 11
As you can see, only v.try_emplace(12, 12, 13) do not use the copy constructor.
both v.emplace(100, Too{ 100,23 }) and v.try_emplace(11, std::move(t)) invoke the copy constructor.
So how can it be even when I use std::move(t)?
Any suggestion would be highly appreciated.
Thanks,
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由于您已经为您的课程提供了一个复制构造函数，因此MOVE构造函数
也:: Too（Too＆amp;＆amp;）Will 不会由编译器隐式生成。此外，如果没有可用于类的移动构造函数，则可以使用复制构造函数。
为了使用移动构造函数，您必须明确提供适当的用户定义的移动构造函数
to :: Too（To to＆amp;＆amp;），那么您将获得所需的结果。您可以通过添加
（又有＆amp;＆amp;）=默认值;或编写自己的移动构造函数来添加MOVE构造函数，该构造构建器将在构造函数初始化器列表中执行初始化。Since you've provided a copy constructor for your class, the move constructor
Too::Too(Too&&)will not be implicitly generated by the compiler.Moreover, when there is no move constructor available for a class, the copy constructor can be used.
For using the move constructor you have to explicitly provide an appropriate user-defined move constructor
Too::Too(Too&&), then you will get the desired result.You can add the move constructor either by adding
Too(Too&&) = default;or writing your own move constructor which will do the initialization in the constructor initializer list.