enable_if to Add a function parameter that has a default argument?

I can't understand the second scenario presented here. It says:

•Scenario 2: Adding a function parameter that has a default argument:

template <your_stuff> your_return_type_if_present
yourfunction(args, enable_if_t<your condition, FOO> = BAR) {
    // ...
}

Scenario 2 leaves the parameter unnamed. You could say ::type Dummy = BAR, but the name Dummy is irrelevant, and giving it a name is likely to trigger an "unreferenced parameter" warning. You have to choose a FOO function parameter type and BAR default argument. You could say int and 0, but then users of your code could accidentally pass to the function an extra integer that would be ignored. Instead, we recommend that you use void ** and either 0 or nullptr because almost nothing is convertible to void **:

template <your_stuff> your_return_type_if_present 
yourfunction(args, typename enable_if<your_condition, void **>::type=nullptr) {
 // ...
}

If scenario 2 leaves the parameter unnamed then in what can it be used? Is there a way to make a code like this work with enable_if?

enum otype {oadd,omull};
template<otype o>
int add(int num1, std::enable_if<o == oadd, int>::type int= num2)
{
    if (o == omull) return num1 * num1;
    if (o == oadd ) return num1 + num2;
 }