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Modern C++ std::move的实现原理

news 2025/7/7 10:04:11

前言

有一节我们分析了std::bind的实现原理，本节稍作休息分析一个比较简单的std::move

原理

std::move的原理太简单了，一句话：就是把传进来的参数强转为右值引用类型。作用为调用移动构造函数，或移动赋值函数。下面通过例子和代码说明。

例子

#include <iostream>
#include <cstring>using namespace std;struct Thing {Thing(){cout << "default construct\n";}// Copy operatorThing(const Thing& other){cout << "copy constructor\n";}// Move constructorThing(Thing&& other) noexcept{cout << "move constructor\n";}// assignment operatorThing& operator=(const Thing& rhs) {cout << "copy operator\n";return *this;}// move assignment operatorThing& operator=(Thing&& rhs) noexcept {cout << "move operator\n";return *this;}// destructor necessary since we are working in dangerous new/delete territory~Thing() noexcept {cout << "destructor " << "\n";}
};
int main()
{cout << "main::constructing a\n";Thing a;cout << "main::moving a to newly constructed c\n";Thing c(std::move(a));Thing x = std::move(c);cout << ">main::thing y\n";Thing y;y = std::move(x);return 0;
}

[mzhai@c++11]$ ./a.out
main::constructing a
default construct
main::moving a to newly constructed c
move constructor //Thing c(std::move(a));
move constructor //Thing x = std::move©;
main::thing y
default construct
move operator //y = std::move(x);
destructor
destructor
destructor
destructor

a, c, x虽然都是左值，但std::move却把它们强转成了右值引用，从而调用了move语义的函数而不是copy语义的。

std::move源码

 92   /**93    *  @brief  Convert a value to an rvalue.94    *  @param  __t  A thing of arbitrary type.95    *  @return The parameter cast to an rvalue-reference to allow moving it.96   */97   template<typename _Tp>98     constexpr typename std::remove_reference<_Tp>::type&&99     move(_Tp&& __t) noexcept
100     { return static_cast<typename std::remove_reference<_Tp>::type&&>(__t); }

让我们看看remove_reference 的定义，看看它是怎么脱去类型的外衣的：

1402   /// remove_reference
1403   template<typename _Tp>
1404     struct remove_reference
1405     { typedef _Tp   type; };
1406
1407   template<typename _Tp>
1408     struct remove_reference<_Tp&>
1409     { typedef _Tp   type; };
1410
1411   template<typename _Tp>
1412     struct remove_reference<_Tp&&>
1413     { typedef _Tp   type; };