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How to use array of pointers to int array in C

3 Answers

0 votes
#include <stdio.h>

#define LEN 10

int main() 
{ 
  int arr[10] = {5, 7, 2, 5, 9, 1, 0, 7, 8, 6};
  int *p[3];
  
  p[0] = arr;
  p[0][0] = 99;
  for (int i = 0; i < LEN; i++)
        printf("%2i", p[0][i]);
  
  
  return 0; 
}



/*
run:

99 7 2 5 9 1 0 7 8 6

*/

 




answered Apr 6 by avibootz
0 votes
#include <stdio.h>

#define LEN 10

int main() 
{ 
  int arr[10] = {5, 7, 2, 5, 9, 1, 0, 7, 8, 6};
  int *p[3];
  
  p[1] = &arr[2];
  p[1][0] = 99;
  for (int i = 0; i < LEN; i++)
        printf("%3i", arr[i]);
  
  
  return 0; 
}



/*
run:

  5  7 99  5  9  1  0  7  8  6

*/

 




answered Apr 6 by avibootz
0 votes
#include <stdio.h>

#define LEN 10

int main() 
{ 
  int arr[10] = {5, 7, 2, 5, 9, 1, 0, 7, 8, 6};
  int *p[3];
  
  p[2] = &arr[6];
  p[2][1] = 99;
  for (int i = 0; i < LEN; i++)
        printf("%3i", arr[i]);
        
  printf("\n%3i\n", p[2][0]);
  printf("%3i\n", p[2][1]);
  printf("%3i\n", *(p[2] + 2));
  
  return 0; 
}



/*
run:

  5  7  2  5  9  1  0 99  8  6
  0
 99
  8

*/

 




answered Apr 6 by avibootz
edited Apr 6 by avibootz
...