Make a combination of two arrays to form a pair of each item

Question

I have two arrays array1 of size 3 and array2 of size 2. I wish to form a pair of each item from both arrays. That is;

int array1[] = {1, 2, 3};
int array2[] = {9, 4};

Results I'm hoping to achieve:
1 , 9
1 , 4
2 , 9
2 , 4
3 , 9
3 , 4

This is what I have tried:

#include <iostream>

using namespace std;

int main(int argc, const char *argv[])
{
    int array1[] = {1, 2, 3};
    int array2[] = {9, 4};

    int arrayOneSize = sizeof(array1);
    int arrayTwoSize = sizeof(array2);

    for (size_t i = 0; i < arrayOneSize; i++)
    {
        for (size_t j = 0; j < arrayTwoSize; j++)
        {
            cout << array1[i] << " , " << array2[j] << endl;
        }

    }

    return 0;
}

But for some reason I am getting a whole bunch of weird combinations like:
1,9
1,4
1,1
1,2
1,3
1,1029505037
1,-531587312
... (It's really long, just want to shorten the results a little)
0,-411331072
1,9
1,4
1,1
1,2
1,3
1,1029505037
1,-531587312
1,-411331072

Sorry for the noob question. I am still new to C++ so, I will gladly appreciate any help. And also why am I getting numbers which are not part of the array?

Thanks in advance.

`int arrayOneSize = sizeof(array1);` ==> `int arrayOneSize = sizeof(array1) / sizeof(*array1);` — mch, Jun 11 '19 at 11:09
https://stackoverflow.com/questions/37538/how-do-i-determine-the-size-of-my-array-in-c — rtoijala, Jun 11 '19 at 11:09
maybe not that obvious, but this question is very similar and the accepted answer shows a generic solution that you could also apply here https://stackoverflow.com/questions/13788114/cross-product-in-c — 463035818_is_not_an_ai, Jun 11 '19 at 11:15

Aykhan Hagverdili · Accepted Answer · 2019-06-11T11:27:59.493

4

To get the number of elements in an array, you have to divide its size by the size of an element in it:

int arrayOneSize = sizeof(array1) / sizeof(array1[0]);
int arrayTwoSize = sizeof(array2) / sizeof(array2[0]);

As Jarod42 points out, you can use the range-based for loop for the arrays and bypass finding the size manually all-together.
Alternatively, consider using std::array, or better yet, std::vector. Here's an implementation with modern C++:

#include <iostream>
#include <vector>

int main() {
  std::vector vec1{ 1, 2, 3 };  // no need for <int> with C++17
  std::vector vec2{ 9, 4 };

  for (auto&& e1 : vec1) {
    for (auto&& e2 : vec2) {
      std::cout << e1 << " , " << e2
                << '\n';  // no need to flush the buffer all the time
    }
  }
}

edited Jun 11 '19 at 11:27

answered Jun 11 '19 at 11:11

Aykhan Hagverdili

28,141
6
41
93

2

BTW, for-range is also available for `int array1[] = {1, 2, 3};` – Jarod42 Jun 11 '19 at 11:25
do you know of any STL algorithm that accomplishes that ? feels like a lot of boiler plate – Midnight Exigent Jun 08 '20 at 00:00
@WalidJabari `std::size` does that. – Aykhan Hagverdili Jun 08 '20 at 00:01
I was refering to the part that generates all the possible permutations – Midnight Exigent Jun 08 '20 at 00:03
@WalidJabari [std::next_permutation](https://en.cppreference.com/w/cpp/algorithm/next_permutation) – Aykhan Hagverdili Jun 08 '20 at 00:04
uh, had that one in mind too, I guess there's no better way than 2 for-loops. thx anyway – Midnight Exigent Jun 08 '20 at 00:07
1

@wal on the second look, next_permutation is not really suitable here. You can always write your own function though. – Aykhan Hagverdili Jun 08 '20 at 00:08

score 2 · Answer 2 · answered Jun 11 '19 at 11:10

The size ( no. of elements in each array) computation of arrays is having the issue.

int arrayOneSize = sizeof(array1);
int arrayTwoSize = sizeof(array2);

This will be number of elements * sizeof(int) which gives the total size. Since you are interested only in number of elements, you should rather do

int arrayOneSize = sizeof(array1)/sizeof(int);
int arrayTwoSize = sizeof(array2)/sizeof(int);

This should give the expected output

Make a combination of two arrays to form a pair of each item

2 Answers2