Template default argument loses its reference type

For foo<int>(a), ARG_T is being deduced from a, and is not taken from the default template argument. Since it’s a by value function parameter, and a is an expression of type int, it’s deduced as int.

In general, default template arguments are not used when template argument deduction can discover what the argument is.

But we can force the use of the default argument by introducing a non-deduced context for the function parameter. For instance:

template <class T, class ARG_T = T&>
T foo(std::enable_if_t<true, ARG_T> v1){
    //...
}

Or the C++20 type_identity utility, such as the other answer demonstrates.