I've created a function to return all the indices where an element is to be found in an array. If it occurs twice or more, all the respective indices will be returned, for now it returns them in natural counting (1,2,3...) and not how the computer sees them (0,1,2,...).

Thanks for your time.

#include <iostream>
#include <vector>
using namespace std;

void searchList(int theArray[], int sizeOfTheArray, int findFor);

void searchList(int theArray[], int sizeOfTheArray, int findFor){
  vector<int> foundIndices;

  int j = 0;

  for (int i = 0; i < sizeOfTheArray; i++)
  {
    if (theArray[i] == findFor){
      foundIndices.push_back(i);
      j++;
    }
  }

  if (foundIndices.size()!=0){
    cout << "Found in index: ";
    for (int i = 0; i < foundIndices.size(); i++){
      cout << foundIndices[i]+1 << " ";
    }
  }
  else
    cout << "Not found in array";
}

int main(){
  int test[8] = {87, 75, 98, 100, 100, 234, 265, 9};
  searchList(test, 8, 99);
  return 0;
}

I've created a function to return all the indices where an element is to be found in an array. If it occurs twice or more, all the respective indices will be returned, for now it returns them in natural counting (1,2,3...) and not how the computer sees them (0,1,2,...).

#include <iostream>
#include <vector>
using namespace std;

void searchList(int theArray[], int sizeOfTheArray, int findFor);

void searchList(int theArray[], int sizeOfTheArray, int findFor){
  vector<int> foundIndices;

  int j = 0;

  for (int i = 0; i < sizeOfTheArray; i++)
  {
    if (theArray[i] == findFor){
      foundIndices.push_back(i);
      j++;
    }
  }

  if (foundIndices.size()!=0){
    cout << "Found in index: ";
    for (int i = 0; i < foundIndices.size(); i++){
      cout << foundIndices[i]+1 << " ";
    }
  }
  else
    cout << "Not found in array";
}

int main(){
  int test[8] = {87, 75, 98, 100, 100, 234, 265, 9};
  searchList(test, 8, 99);
  return 0;
}