Do expressions fun and &fun have the same type or not?
Consider the following code:
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
Why is that?
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);
fun and &fun refer to the same type because of function to pointer conversion, which is performed in check<void(*)()>(fun);; but typeid is an exception.
(emphasis mine)
Lvalue-to-rvalue, array-to-pointer, or function-to-pointer conversions are not performed.
And why the function to pointer conversion is performed for check<void(*)()>(fun);, because in template argument deduction,
Before deduction begins, the following adjustments to P and A are made:
1) If P is not a reference type,
- if A is an array type, ...;
- otherwise, if A is a function type, A is replaced by the pointer type obtained from function-to-pointer conversion;
check() takes parameter by value, then function-to-pointer conversion is performed and the deduced type of T will be the function pointer too, i.e. void(*)().
When you use a function name as an expression, it decays to a pointer to itself. So fun will be the same as &fun.
As for the typeid thing, from this reference:
Lvalue-to-rvalue, array-to-pointer, or function-to-pointer conversions are not performed.
[Emphasis mine]
