CODEWITHSUNDEEP
carrayspointers
fishy
fishy

Reputation: 3

Using *only* pointers to print the contents of an array

I have been asked to write a function that prints the content of an array after the xth element (as in, x and forth, as x is a pointer inside the array). I am not allowed to use [] for anything other than initialization, and not allowed to create variables inside the function - I may only use what the function receives from the main, which is the length (int n), the array (int* arr) and element x (int* x).

My question is how can I print x and forth in the array using only pointers without an index from a loop?

Here is what I wrote:

void printAfterX(int* arr, int n, int* x)
{
    if ((arr <= x) && (x < arr + n))
    {
        while(x < (arr + n))
        {
            printf("%8d", *(arr+x));        //I know you can't do this
            x++;
        }
    }
}

For this: 

    int arr[] = { 0,5,6,7,8,4,3,6,1,2 };
    int n=10;
    int* x = (arr+3);

    printAfterX(arr, n, x);

The output should be:

7 8 4 3 6 1 2

Edit: Thanks y'all for the help! Works just fine. :)

Upvotes: 0

Views: 238

Answers (3)

Jabberwocky
Jabberwocky

Reputation: 50778

You want this:

#include <stdio.h>
#include <stdlib.h>

void printAfterX(int* arr, int n, int* x)
{
  while(x < (arr + n))
  {
      printf("%d ", *x);
      x++;
  }
}

int main()
{
  int arr[] = { 0,5,6,7,8,4,3,6,1,2 };
  int n = 10;
  int *x = (arr+3);

  printAfterX(arr, n, x);
  return 0;
}

Upvotes: 0

CiaPan
CiaPan

Reputation: 9571

void printAfterX(int* arr, int n, int* x)
{
    arr += n;               // make arr to point past the last element of the array
    for( ; x < arr; x++)    // walk till the end of array
        printf("%8d", *x);  // print the current item
}

Example https://ideone.com/Ea3ceT

Upvotes: 2

atturri
atturri

Reputation: 1153

Change the printf line for this:

printf("%8d", *x);

Upvotes: -1

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