Do expressions fun and &fun have the same type or not?
考虑以下代码：
template <typename Check, typename T>
void check(T)
{
static_assert(is_same<Check, T>::value);
}
void fun()
{}
check<void(*)()>(fun);
check<void(*)()>(&fun);
cout << typeid(fun).name() << endl;
cout << typeid(&fun).name() << endl;
Both assertions succeed which suggests that both expressions have the same type. However, typeids return different results:
FvvE
PFvvE
这是为什么？
最佳答案
Both assertions succeed because they are applied to the type T deduced from function argument. In both cases it will be deduced as a pointer to function because functions decay to a pointer to function. However if you rewrite assertions to accept types directly then first one will fail:
static_assert(is_same<void(*)(), decltype(fun)>::value);
static_assert(is_same<void(*)(), decltype(&fun)>::value);
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