Template deduction interesting case, c++ -

consider peace of code:

template<class t> void f(const t& t) {     static int x = 0;     cout<<++x<<endl; }  int main() {     int j = 0;     const int = 0;     f(5);     f(i);     f(j); } 

i have called function 3 types. although 5 , j can same thing, int, const int different type.
anyway output is:

1 2 3 

so means compiler instantiates same function different types.
correct? can explain why?

from [temp.deduct.call]:

template argument deduction done comparing each function template parameter type (call p) type of corresponding argument of call (call a) described below.

p const t& , a int, int, , const int in 3 calls.

we have:

if p reference type, type referred p used type deduction.

p reference type, use p' == const t deduction against a == int or a == const int. in both cases, deduce t == int p' == const int (and p == const int&) , deduced a == const int. more cv-qualified original a first 2 calls, that's explicitly made ok:

in general, deduction process attempts find template argument values make deduced identical (after type transformed described above). however, there 3 cases allow difference:
— if original p reference type, deduced (i.e., type referred reference) can more cv-qualified transformed a.

thus, 3 cases call f<int>.