In the code below, why do the two ways of invoking fun
: fun(num)
and fun<const int>(num)
, give a different result when compiling?
#include <iostream>
using namespace std;
template<typename T, typename = typename enable_if<!std::is_same<int, T>::value>::type>
void fun(T val)
{
cout << val << endl;
}
int main(void)
{
const int num = 42;
fun(num); //ERROR!
fun<const int>(num); //Right
return 0;
}
The parameter is declared as pass-by-value; then in template argument deduction, the top-level const qualifier of the argument is ignored.
Before deduction begins, the following adjustments to P and A are made:
1) If P is not a reference type,
a) ...
b) ...
c) otherwise, if A is a cv-qualified type, the top-level cv-qualifiers are ignored for deduction:
So given fun(num)
, the template parameter T
will be deduced as int
, not const int
.
If you change the parameter to pass-by-reference, the const
part will be preserved. e.g.
template<typename T, typename = typename enable_if<!std::is_same<int, T>::value>::type>
void fun(T& val)
Then for fun(num)
, T
will be deduced as const int
.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With