Question

我想知道为什么std :: vector在动态增长时不使用move构造函数。或者我的代码有问题吗？请参阅下面的测试代码：

#include<vector>
#include<iostream>
#include<cstring>

using namespace std;
class test{
public:
    char *ptr; 
public:
    test()
        :ptr{new char[10]}{
            strcpy(ptr,"hello");
            cout<<"constructor called"<<endl;
        }
    test(const test& t){
        ptr = new char[10];
        strcpy(ptr,t.ptr);
        cout<<"copy constructor called"<<endl;
    }
    test(test &&t){
        this->ptr = t.ptr;
        t.ptr = nullptr;
        cout<<"move constructor called"<<endl;
    }
    ~test(){
        cout<<"destructor called"<<endl;
        delete[] ptr;
    }
};

vector<test> function()
{
    vector<test> v;
    cout<<v.size()<<" "<<v.capacity()<<endl;
    v.push_back(test{});
    cout<<v.size()<<" "<<v.capacity()<<endl;
    v.push_back(test{});
    cout<<v.size()<<" "<<v.capacity()<<endl;
    return v;
}

int main()
{ 
    vector<test> v = function();
}

上述代码的输出如下：

0 0
constructor called
move constructor called
destructor called
1 1
constructor called
move constructor called
copy constructor called
destructor called
destructor called
2 2
destructor called
destructor called

即使类test具有移动构造函数，仍然向量使用复制构造函数在内部调整大小时。这是预期的行为还是它因为test类本身存在一些问题？

感谢您的回答。

当vector动态增长时，为什么它不使用move构造函数？

0 个答案: