停用std::forward 的範本參數推導以確保正確轉送
考慮VS2010 中std::forward 的定義：
<code class="cpp">template<class _Ty> inline
_Ty&& forward(typename identity<_Ty>::type& _Arg)
{ // forward _Arg, given explicitly specified type parameter
return ((_Ty&&)_Arg);
}</code><code class="cpp">template<typename T>
T&& forward_with_deduction(T&& obj)
{
return static_cast<T&&>(obj);
}</code><code class="cpp">void test(int&){}
void test(const int&){}
void test(int&&){}
template<typename T>
void perfect_forwarder(T&& obj)
{
test(forward_with_deduction(obj));
}
int main()
{
int x;
const int& y(x);
int&& z = std::move(x);
test(forward_with_deduction(7)); // 7 is an int&&, correctly calls test(int&&)
test(forward_with_deduction(z)); // z is treated as an int&, calls test(int&)
// All the below call test(int&) or test(const int&) because in perfect_forwarder 'obj' is treated as
// an int& or const int& (because it is named) so T in forward_with_deduction is deduced as int&
// or const int&. The T&& in static_cast<T&&>(obj) then collapses to int& or const int& - which is not what
// we want in the bottom two cases.
perfect_forwarder(x);
perfect_forwarder(y);
perfect_forwarder(std::move(x));
perfect_forwarder(std::move(y));
}</code>以上是為什麼 std::forward 使用恆等模板來停用模板參數推導？的詳細內容。更多資訊請關注PHP中文網其他相關文章！
