Why does my function print an address instead of the contents of my array?

Question

My code is supposed to remove a specified value and shift the array. When I run the code it prints an address instead of the contents of the array. What is wrong with my code?

using namespace std;
void arrayShift(int arr[], int length, int value) {
  for(int i = 0; i<length; i++) {
    if(arr[i] == value) {
      for (int k = i; k<length-1 ; k++) {
        arr[k] = arr[k+1];
      }
      arr[length-1] = 0;
      i--;
    }
  }
  cout << arr;
}

int main() {
  int inputarr[]={9,8, 9, 9, 9, 9, 6};
  int length = 7;
  int value = 9;
  arrayShift(inputarr,length,value);
}

It's not clear what you are trying to do. You can't expect your function to return your array if you specified void... — AMDG, Mar 13 '15 at 15:19
a `void` function is one that returns nothing. so... which word did you mean to use instead of *return*? — just somebody, Mar 13 '15 at 15:21
`std::remove(std::begin(inputarr), std::end(inputarr), value);` — Neil Kirk, Mar 13 '15 at 15:23
You mean, why does `cout << arr` print the address, not the whole array? Because that's what `< — Mike Seymour, Mar 13 '15 at 15:24
@Andrew If I understand your question, a more accurate title would be "Why does my function print...". This doesn't seem to be a question about returning or return types. — Drew Dormann, Mar 13 '15 at 15:30
possible duplicate of [Printing an array in C++?](http://stackoverflow.com/questions/1370323/printing-an-array-in-c) — sashoalm, Mar 13 '15 at 15:34
If my response answer your wuestion, accept it as answer by checking the green mark. Thanks ! — AMDG, Mar 13 '15 at 15:42
@sashoalm almost: but I think the "why doesn't `cout << arr` work?" isn't answered by the answers there, as the asker already knew that. — Yakk - Adam Nevraumont, Mar 13 '15 at 16:23
@Yakk Yes, I was a bit more flexible in my interpretation of exact duplicate. My idea was that a 'redirect' to that question was the most sensible thing for future visitors. — sashoalm, Mar 13 '15 at 16:26

score 3 · Accepted Answer · answered Mar 13 '15 at 15:24

3

The line

cout << arr;

Display an address because cout doesn't display array directly.

You should use a for to display all your values. Something like that:

for(int i = 0 ; i < length ; i++)
    cout << arr[i];

However you really should put the output (for with cout) in the main function.

answered Mar 13 '15 at 15:24

AMDG

945
1
9
22

My code will still print the address when I make the changes. – Andrew Mar 13 '15 at 15:50
cpp.sh/7rbv There seems to be an error with the way I am calling it. Why cannot I not just cout my function? – Andrew Mar 13 '15 at 15:57
Your function return void, i.e. nothing. cout will try to display what your function will return, so cout will try to display nothing. – AMDG Mar 13 '15 at 16:00
What would be the correct way of calling the function then? – Andrew Mar 13 '15 at 16:02
The correct way is to display your array, not what your function return. See : http://cpp.sh/7mld – AMDG Mar 13 '15 at 16:03

Why does my function print an address instead of the contents of my array?

1 Answers1