C++ utility forward用法及代碼示例

描述

如果 arg 不是左值引用，則它返回對 arg 的右值引用。

聲明

以下是 std::forward 函數的聲明。

template <class T> T&& forward (typename remove_reference<T>::type& arg) noexcept;
template <class T> T&& forward (typename remove_reference<T>::type&& arg) noexcept;

C++11

template <class T> T&& forward (typename remove_reference<T>::type& arg) noexcept;
template <class T> T&& forward (typename remove_reference<T>::type&& arg) noexcept;

參數

arg- 它是一個對象。

返回值

如果 arg 不是左值引用，則它返回對 arg 的右值引用。

異常

Basic guarantee- 該函數從不拋出異常。

數據競爭

空

示例

在下麵的示例中解釋了 std::forward 函數。

#include <utility>
#include <iostream>

void overloaded (const int& x) {std::cout << "[It is a lvalue]";}
void overloaded (int&& x) {std::cout << "[It is a rvalue]";}

template <class T> void fn (T&& x) {
   overloaded (x);
   overloaded (std::forward<T>(x));
}

int main () {
   int a;

   std::cout << "calling fn with lvalue:";
   fn (a);
   std::cout << '\n';

   std::cout << "calling fn with rvalue:";
   fn (0);
   std::cout << '\n';

   return 0;
}

讓我們編譯並運行上麵的程序，這將產生以下結果 -

calling fn with lvalue:[It is a lvalue][It is a lvalue]
calling fn with rvalue:[It is a lvalue][It is a rvalue]