Print spaces between each element using a fold expression

Question

I am using a fold expression to print elements in a variadic pack, but how do I get a space in between each element?

Currently the output is "1 234", the desired output is "1 2 3 4"

template<typename T, typename Comp = std::less<T> >
struct Facility
{
template<T ... list>
struct List
{
    static void print()
    {

    }
};
template<T head,T ... list>
struct List<head,list...>
{
    static void print()
    {
     std::cout<<"\""<<head<<" ";
     (std::cout<<...<<list);
    }
};
};

template<int ... intlist>
using IntList = typename Facility<int>::List<intlist...>;
int main()
{
 using List1 = IntList<1,2,3,4>;
 List1::print();
}

Sorry!, i edited the code, should compile now – Lim Ta Sheng Jul 11 '18 at 09:32 — Lim Ta Sheng, Jul 11 '18 at 09:32

Tyker · Accepted Answer · 2018-07-11T11:03:38.993

you can that

#include <iostream>

template<typename T>
struct Facility
{
template<T head,T ... list>
struct List
{
    static void print()
    {
     std::cout<<"\"" << head;
     ((std::cout << " " << list), ...);
      std::cout<<"\"";
    }
};
};

template<int ... intlist>
using IntList = typename Facility<int>::List<intlist...>;
int main()
{
 using List1 = IntList<1,2,3,4>;
 List1::print();
}

the fold expression ((std::cout << " " << list), ...) will expands to ((std::cout << " " << list1), (std::cout << " " << list2), (std::cout << " " << list3)...)

Jarod42 · Answer 2 · 2021-04-14T08:01:57.487

8

If you need space only between numbers (and not after the last or before the first too), you might do:

template <std::size_t... Is>
void print_seq(std::index_sequence<Is...>)
{
    const char* sep = "";
    (((std::cout << sep << Is), sep = " "), ...);
}

Demo

(It is similar to my "runtime version") for regular containers with for-loop.

edited Apr 14 '21 at 08:01

answered Apr 10 '19 at 08:48

Jarod42

190,553
13
166
271

score 7 · Answer 3 · answered Jul 11 '18 at 09:36

7

In general, you use recursion for tasks like this.

You have to define what happens when there are 2 or more and 1 elements in the list and recursively fall back to those definitions:

template <int ...> struct List;
template <int First, int Second, int ... More> struct List {
    static void print() {
        std::cout << First << " ";
        List<Second, More ...>::print();
    }
};
template <int Last> struct List {
    static void print() {
        std::cout << Last;
    }
};

answered Jul 11 '18 at 09:36

Markus Mayr

3,868
1
18
40

but if my packs have thousand of elements that recursion would fail. – Lim Ta Sheng Jul 11 '18 at 09:38
4

Packs might not be the best choice for having thousands of elements, though – deW1 Jul 11 '18 at 09:41
Recursion does `O(N)` instantiations, whereas folding expression allows `O(1)` instantiation (which shouldn't change runtime execution, but compilation time). – Jarod42 Apr 10 '19 at 08:52

Davide Spataro · Answer 4 · 2018-07-11T09:56:26.927

You can reuse print() to achieve this behaviour. Afterall you are doing a fold operation which is by definition resursive.

Live Demo

template<T head,T ... rest_of_pack>

struct List<head , rest_of_pack...>
{
    static void print_()
    {
     std::cout<<head<<" ";
     List<rest_of_pack...>::print();

    }
};

If you want to process many elements this way you might run into problems with template depth (gcc for instance has a limit of 900). Lucky for you you can use the -ftemplate-depth= option to tweak this behaviour.

You can compile with -ftemplate-depth=100000 and make it work. Note that compilation time will skyrocket (most likely) or in thhe worst case you run out of memory.

Print spaces between each element using a fold expression

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