How to expand the initializer list parameters pack?

Question

Codes like:

template <typename... type>
void print(type... pack) {
    ((std::cout << pack << " "), ...);
}

But I have parameters like: { {1, 2, 3}, {4, 5, 6} }

So how can I pass this to the function ? Or, how to expand the parameters pack like this ?

I don't think you can unless you change the function parameter to nested initializer_lsit — Nimrod, Feb 22 '22 at 07:54
`{1, 2, 3}` has no types, and can only be deduced in limited context (`std::initializer_list` or `T(&)[N]`). — Jarod42, Feb 22 '22 at 09:11

score 3 · Answer 1 · answered Feb 22 '22 at 08:21

I suppose you want to expand all the initlizer_list in a row.

// for single initializer_list
template <typename T>
void print(std::initializer_list<T> args) {
    for (const auto arg : args) {
        std::cout << arg << " ";
    }
}

// for nested initializer_list
template <typename T>
void print(std::initializer_list<std::initializer_list<T>> args) {
    for (const auto arg : args) {
        print(arg);
    }
}

See demo

score 3 · Accepted Answer · answered Feb 22 '22 at 08:29

But I have parameters like: { {1, 2, 3}, {4, 5, 6} }

You can pack them into std::tuples

#include<iostream>
#include<tuple>

template <typename... Tuples>
void print_tuples(Tuples... tuples) {
  (std::apply([](auto... args) {
    ((std::cout << args << " "), ...);
  }, tuples), ...);
}

Then

print_tuples(std::tuple{1, 2, 3}, std::tuple{4, 5, 6});

Demo

How to expand the initializer list parameters pack?

2 Answers2