Array pointers differences in two types of declarations

Question

Is anything wrong with the following piece of code?

int x[3][12];
int *ptr[12];
ptr=x;

This isn't exactly the typical format for StackOverflow questions. Typically, questions follow this pattern: "This is my code. This is the problem. I've tried this solution and it didn't work. How can I fix this problem?" — nhgrif, Dec 12 '13 at 23:05

mclaassen · Answer 1 · 2013-12-12T23:38:47.263

Yes, it fails to compile with the message:

'=' : cannot convert from 'int [3][12]' to 'int *[12]'

Explanation:

Here x is a 2 dimensional array of ints, while ptr is an array of pointers to int.

The statement ptr=x is trying to assign x (which is a 2 dimensional array of ints) to ptr (which is an array of pointers to int).

If you were to declare ptr as

int (*ptr)[12];

instead, then ptr would be a pointer to an array of ints, and since a pointer to an array of ints is essentially the same thing as a 2 dimensional array of ints (an array of ints is just a pointer to the first element in the array, and a 2 dimensional array of ints is just a pointer to the first array of ints) then it would compile.

Alternatively consider this:

int x[3][12];
int *ptr[12];
ptr[0] = x[0];

which is a valid assignment of an array of ints to a pointer to int (note that the fact that the second dimension of x is the same as ptr is not why this works)

You might wanna add that `int (*ptr)[12]` will make the code valid though. — JustinBlaber, Dec 12 '13 at 23:14

score 1 · Answer 2 · edited May 23 '17 at 10:32

Have a look at: C pointer to array/array of pointers disambiguation

I got a bit intrigued by this and did a working example:

#include <stdio.h>

int main (int argc, char * argv[])
{
  int i,j;
  int x[3][12];    // An array of arrays with lenght 12
//int *wrong[12];  // An array of pointers to ints.
  int (*ptr)[12];  // A pointer to an array of ints

  /* Initialization to consequtive numbers*/
  for ( i = 0; i < 12*3; i++)
  {
    /* Get the first index, and go from there */
    int *number = &(x[0][0])+i;
    *number = i;
  }


  /* Iterate throug and check that the pointers are the same */
  for ( i = 0; i < 3 ; i ++ )
  {
    /* This will generate a warning: 
     * assignment from incompatible pointer type [enabled by default]
     */
    ptr = x[i];

    fprintf(stderr,"Numbers in ptr: ");
    for ( j = 0; j < 12; j ++)
      fprintf(stderr,"%2d ", (*ptr)[j]);
    fprintf(stderr,"\n");

    fprintf(stderr,"Numbers in x:   ");
    for ( j = 0; j < 12; j ++)
      fprintf(stderr,"%2d ", x[i][j]);
    fprintf(stderr,"\n");
  }

  return 0;
}

The output is:

Numbers in ptr:  0  1  2  3  4  5  6  7  8  9 10 11 
Numbers in x:    0  1  2  3  4  5  6  7  8  9 10 11 
Numbers in ptr: 12 13 14 15 16 17 18 19 20 21 22 23 
Numbers in x:   12 13 14 15 16 17 18 19 20 21 22 23 
Numbers in ptr: 24 25 26 27 28 29 30 31 32 33 34 35 
Numbers in x:   24 25 26 27 28 29 30 31 32 33 34 35

Array pointers differences in two types of declarations

2 Answers2