Tuple to parameter pack

Question

This below code from user Faheem Mitha, is based on user Johannes Schaub - litb\'s answer in this SO. This code perfectly does what I seek, which is conversion of a tu

Answer 1

Let's look at what happens here:

template struct gens : gens { };

template struct gens<0, S...>{ typedef seq type; };

The first one is a generic template, the second one is a specialization that applies when the first template parameter is 0.

Now, take a piece of paper and pencil, and write down how

 gens<3>

gets defined by the above template. If your answer was:

 struct gens<3> : public gens<2, 2>

then you were right. That's how the first template gets expanded when N is "3", and ...S is empty. gens, therefore, becomes gens<2, 2>.

Now, let's keep going, and see how gens<2, 2> gets defined:

 struct gens<2, 2> : public gens<1, 1, 2>

Here, in the template expansion, N is 2, and ...S is "2". Now, let's take the next step, and see how gens<1, 1, 2> is defined:

 struct gens<1, 1, 2> : public gens<0, 0, 1, 2>

Ok, now how does gens<0, 0, 1, 2> gets defined? It can now be defined by the specialization:

 template struct gens<0, S...>{ typedef seq type; };

So, what happens with struct gens<0, 0, 1, 2> here? Well, in the specialization, "S..." becomes "0, 1, 2", so this becomes, in a manner of speaking:

 struct gens<0, 0, 1, 2> {

   typedef seq<0, 1, 2> type;

 }

Now, keep in mind that all of these publicly inherit from each other, "elephant-style", so:

 gens<3>::type

ends up being a typedef declaration for

 struct seq<0, 1, 2>

And this is used, by the code that follows to convert the tuple into a parameter pack, using another template:

double delayed_dispatch()
{
    return callFunc(typename gens::type()); // Item #1
}

...Args are the tuple parameters. So, if there are three elements in the tuple, sizeof(...Args) is 3, and as I've explained above, gens::type() becomes gens<3>::type(), a.k.a. seq<0, 1, 2>().

So, now:

template
double callFunc(seq)
{
    return func(std::get(params) ...);
}

The S... part becomes "0, 1, 2", so the

std::get(params)...

Becomes a parameter pack that gets expanded to:

std::get<0>(params), std::get<1>(params), std::get<2>(params),

And that's how a tuple becomes a parameter pack.