c++ - read arguments from variadic template

Question

I am a little confused about how can I read each argument from the tuple by using variadic templates.

Consider this function:

template<class...A> int func(A...args){
int size = sizeof...(A);
.... }

I call it from the main file like:

func(1,10,100,1000);

Now, I don't know how I have to extend the body of func to be able to read each argument separately so that I can, for example, store the arguments in an array.

Answer 1

You have to provide overrides for the functions for consuming the first N (usually one) arguments.

void foo() {
   // end condition argument pack is empty
}

template <class First, class... Rest> 
void foo(First first, Rest... rest) {
    // Do something with first
    cout << first << endl; 

    foo(rest...); // Unpack the arguments for further treatment
}

When you unpack the variadic parameter it finds the next overload.

Example:

foo(42, true, 'a', "hello");
// Calls foo with First = int, and Rest = { bool, char, char* }
// foo(42, Rest = {true, 'a', "hello"}); // not the real syntax

Then next level down we expand the previous Rest and get:

foo(true, Rest = { 'a', "hello"}); // First = bool

And so on until Rest contains no members in which case unpacking it calls foo() (the overload with no arguments).

---

Storing the pack if different types

If you want to store the entire argument pack you can use an std::tuple

template <class... Pack>
void store_pack(Pack... p) {
    std::tuple<Pack...> store( p... );
    // do something with store
}

However this seems less useful.

Storing the pack if it's homogeneous

If all the values in the pack are the same type you can store them all like this:

vector<int> reverse(int i) {
    vector<int> ret;
    ret.push_back(i);
    return ret;
}

template <class... R>
vector<int> reverse(int i, R... r) {
    vector<int> ret = reverse(r...);
    ret.push_back(i);
    return ret; 
}

int main() {
    auto v = reverse(1, 2, 3, 4);
    for_each(v.cbegin(), v.cend(), 
        [](int i ) { 
            std::cout << i << std::endl; 
        }
    );
}

However this seems even less useful.