问题描述
class string
{
std::string data;
public:
string()
{
data = "Hello";
}
string(const string& that)
{
data = std::move(that.data);
std::cout << data << " and that.data " << that.data;
}
};
int main()
{
std::vector<int> vec = { 1,2,3,4,5 };
std::cout << "vec size = " << vec.size() << std::endl;
std::vector<int> nvec = std::move(vec);
std::cout << "nvec size = " << nvec.size() << std::endl;
std::cout << "vec size = " << vec.size() << std::endl;
string s;
string s1(s);
}
vec size = 5
nvec size = 5
vec size = 0
Hello and that.data Hello
为什么 that.data 没有像 nvec 和 vec 那样移动到目标数据?
解决方法
您将参数 that
声明为 const
。因此,data = std::move(that.data);
调用复制赋值运算符,而不是移动赋值运算符。移动操作应该是对要移动的对象进行修改,而不应该是const
。
您可以将参数类型更改为右值引用(到非常量)。
string(string&& that)
{
data = std::move(that.data);
std::cout << data << " and that.data " << that.data;
}
然后
string s;
string s1(std::move(s));