Template non-type parameter deduction

Question

Is it possible to deduce template value (not type) for a c++17 function?

The function foo:

template<int I>
int foo()
{
    return (I);
}

Can be called via:

foo<5>();

And will return 5.

Template types can be deduced through the type of a function argument. Is it possible to do the same in some way for the template value? For example:

template<int I = x>
int bar(const int x)
{
    return (I);
}

This obviously wont work (because for one x is required before its declaration), but might there be some C++17 trick which would allow for this?

I'd like to use this as a way of setting a constant expression function argument.

Answer 1

What you want can only be done by (ab)using type deduction for integer deduction. Observe:

template<int x>
struct integer_value {};

template<int x>
void test(integer_value<x> val)
{
  //x can be used here.
}

Of course, you must invoke this with test(integer_value<4>{}) or something similar.