我想知道C++中的move函数的内部结构。为此,我打算自己实现名为move2的move函数。
下面是我的实现,通过打印跟踪内存分配。
#include <iostream>
#include <vector>
using namespace std;
void * operator new(size_t size) {
cout << "New operator overloading " << endl;
void * p = malloc(size);
return p;
}
void operator delete(void * p) {
cout << "Delete operator overloading " << endl;
free(p);
}
template<typename T>
T move2(T& input) {
cout << "Running move2 " << endl;
return (T&&)input;
}
int main()
{
{
cout<<"Test#1"<<endl;
vector<int> temp1(10,4);
vector<int> temp2 = temp1;
}
{
cout<<"Test#2"<<endl;
vector<int> temp3(10,4);
vector<int> temp4 = (vector<int>&&)(temp3);
}
{
cout<<"Test#3"<<endl;
vector<int> temp5(10,4);
vector<int> temp6 = move(temp5);
}
{
cout<<"Test#4"<<endl;
vector<int> temp7(10,4);
vector<int> temp8 = move2(temp7);
}
return 0;
}这是输出
Test#1
New operator overloading
New operator overloading
Delete operator overloading
Delete operator overloading
Test#2
New operator overloading
Delete operator overloading
Test#3
New operator overloading
Delete operator overloading
Test#4
New operator overloading
Running move2
Delete operator overloading我想知道我对move2的实现是否正确,我能在生产中使用它吗?
更新:
我发现GCC实现了移动
/**
* @brief Convert a value to an rvalue.
* @param __t A thing of arbitrary type.
* @return The parameter cast to an rvalue-reference to allow moving it.
*/
template<typename _Tp>
constexpr typename std::remove_reference<_Tp>::type&&
move(_Tp&& __t) noexcept
{ return static_cast<typename std::remove_reference<_Tp>::type&&>(__t); }发布于 2021-11-05 05:34:29
std::move()接收一个,并返回一个T&& rvalue引用。你的move2()也不知道。它接受一个lvalue引用,并按值返回一个T。
您的代码“工作”是因为复制省略,避免在函数按值返回新对象时创建临时对象,从而允许直接使用对temp8的类型强制引用构造input,从而转移对其数据的所有权。不过,这并不是因为有了适当的move2()实现。
试一试:
template<typename T>
std::remove_reference_t<T>&& move2(T&& input) {
...
return static_cast<std::remove_reference_t<T>&&>(input);
}https://stackoverflow.com/questions/69848787
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