C++ move semantic移动语义介绍

class MoveClass
{
public:
	int* p;
	MoveClass()
	{
		p = new int();
		std::cout << "默认构造调用" << std::endl;
	}
	~MoveClass()
	{
		std::cout << "析构函数调用" << std::endl;
		if (!p)
			delete p;
	}
	MoveClass(MoveClass& tmp)
	{
 
	}
	 MoveClass(MoveClass&& tmp)
	{
		std::cout << "移动构造函数调用" << std::endl;
		this->p = tmp.p;
		tmp.p = nullptr;
	}
	MoveClass& operator=(MoveClass&& tmp)
	 {
		 std::cout << "移动构造函数调用" << std::endl;
		 this->p = tmp.p;
		 tmp.p = nullptr;
	 }
};
MoveClass MoveClassTest(MoveClass d)
{
	return MoveClass();
}
int main()
{
	MoveClass cc;
	//好 接下来我们不再想使用c了 但是堆区的资源我们并不想拷贝 那么使用如下构造方式
	MoveClass d(std::move(cc));
	system("pause");
}

#include <iostream>
#include <string.h>
void test1(std::string s)
{
	std::cout << "test1:" << s.c_str()<<std::endl;;
}
void test()
{
	std::string s = "超级大的string";
	test1(s);
	std::cout <<"test:"<< s.c_str() << std::endl;
	return;
}
int main()
{
	test();
	system("pause");
}

#include <iostream>
#include <string.h>
void test1(std::string s)
{
	std::cout << "test1:" << s.c_str()<<std::endl;;
}
void test()
{
	std::string s = "超级大的string";
	test1(std::move(s));
	std::cout <<"test:"<< s.c_str() << std::endl;
	return;
}
int main()
{
	test();
	system("pause");
}