场景:
- 在 remove_reference 结构体中能看到右值引用的身影 &&, 那么这里的右值引用究竟有什么用呢?
- 常常也发现int& 和int&& 这两种相似的应用类型,第一种是左值引用声明,另外一种是右值引用声明.
- 下面的样例直接參考的 cppreference,已经写的非常完整了,不是必需又一次写.
參考:
1.Reference declaration
2.Rvalue Reference Declarator: &&
3.Lvalue Reference Declarator: &
语法
左值引用声明: & attr(optional) declarator (1)
右值引用声明: && attr(optional) declarator (2) (since C++11)
左值引用声明 &
–1. 左值引用能够理解为对象的别名, 左值引用必须被初始化并且不能被又一次赋值.
#include <iostream>
#include <string>
int main()
{
std::string s = "Ex";
std::string& r1 = s;
const std::string& r2 = s;
r1 += "ample"; // modifies s
// r2 += "!"; // error: cannot modify through reference to const
std::cout << r2 << '
'; // prints s, which now holds "Example"
}
–2. 在调用函数时,能够被用来作为引用參数.
#include <iostream>
#include <string>
void double_string(std::string& s)
{
s += s; // 's' is the same object as main()'s 'str'
}
int main()
{
std::string str = "Test";
double_string(str);
std::cout << str << '
';
}
–3. 当一个函数的返回值是左值时, 这个函数的调用的表达式能够被觉得是一个左值表达式.
#include <iostream>
#include <string>
char& char_number(std::string& s, std::size_t n)
{
return s.at(n); // string::at() returns a reference to char
}
int main()
{
std::string str = "Test";
char_number(str, 1) = 'a'; // the function call is lvalue, can be assigned to
std::cout << str << '
';
}
右值引用声明 &&
–1. 右值引用声明能够差别一个左值和右值.
–2. 右值引用直至Move语义的实现, 它能够把暂时对象(一般不能引用)的资源移动到其它地方去.
–3. 右值引用也能够用来扩展暂时对象的生命周期,
#include <iostream>
#include <string>
int main()
{
std::string s1 = "Test";
// std::string&& r1 = s1; // error: can't bind to lvalue
const std::string& r2 = s1 + s1; // okay: lvalue reference to const extends lifetime
// r2 += "Test"; // error: can't modify through reference to const
std::string&& r3 = s1 + s1; // okay: rvalue reference extends lifetime
r3 += "Test"; // okay: can modify through reference to non-const
std::cout << r3 << '
';
}
–4. 假设函数有左值和右值引用的重载,调用时会匹配调用, 传左值调用左值重载, 右值调用右值重载. 这样就能够支持Move Constructor 和 Move Assignment 的实现.
#include <iostream>
#include <utility>
void f(int& x)
{
std::cout << "lvalue reference overload f(" << x << ")
";
}
void f(const int& x)
{
std::cout << "lvalue reference to const overload f(" << x << ")
";
}
void f(int&& x)
{
std::cout << "rvalue reference overload f(" << x << ")
";
}
int main()
{
int i = 1;
const int ci = 2;
f(i); // calls f(int&)
f(ci); // calls f(const int&)
f(3); // calls f(int&&)
// would call f(const int&) if f(int&&) overload wasn't provided
f(std::move(i)); // calls f(int&&)
}