完美无缺。 但是,如果我想要而不是<>,那么我怎么做,那么,就什么时候,什么是模版?
template<typename T> void foo(T&& ref);
template<typename T> void foo(const T& ref);
Won t work because the first overload will invoke perfect forwarding. I d really like behaviour very similar to if the first was a normal lvalue reference- where T will always be a value type.
#include <type_traits>
template<typename T>
typename std::enable_if
<
!std::is_reference<T>::value,
void
>::type
foo(T&& ref) {}
template<typename T> void foo(const T& ref) {}
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