Why are char[] and char* as typedefs different, but sometimes… not?

Question

The following observation arose as I was following this question about char[] and char* differences.

#include 

typ         


        

Answer 1

I changed the code so that we could see how calling a f2 changes the type. Before the call the variables are of different type. After the call they have become same

    typedef char ar[];
typedef char* pr;
void f2(ar x, pr y)
{
    cout << is_same::value << '\n'; //same type
}

int main()
{
    ar data = "data";
    pr ptr = data;
    cout << is_same::value << '\n'; // different
    f2(data,ptr);
    return 0;
}

the output is 0 0 .As @jthill, @Dyp and @keith Thompson says this is because of decaying of the array to pointer.