Multidimensional array and pointer addressing

Question

#include <stdio.h>
int main(void)
{
  int x[2][3] =
  {
    {4, 5, 2},
    {7, 6, 9}
  };
  int (*p)[3] = &x[1];
  int (*q)[3] = x;
  printf("%d %d %d\n", (*p)[0], (*p)[1], (*p)[2]);  //7, 6, 9
  printf("%d %d\n", *q[0], *q[1]);  // 4, 7
  return 0;
}

x[0] ----> [4, 5, 2].

x[1] ----> [7, 6, 9]

so if p=X[1], p[0]=7, p[1]=6 and p[2]=9, so the first printf is understandable.

For the second printf, x will be equal to the address first element of the array. If *q[0] is 4, why is *q[1] 7, shouldn't it be 5? It skips a row.

Actual output from link:

7 6 9
4 7

Shouldn't the output of p be 7, 6 and 9, since p points to the second array in x? Please edit the question to include the actual output (copied and pasted) from the actual program you show in the question. — Some programmer dude
– Some programmer dude, Commented Oct 20, 2015 at 5:29
you are correct, sorry fixed. Typed wrong row, but question stands. — chefcurry7
– chefcurry7, Commented Oct 20, 2015 at 5:33
It's a question about operator precedence, the array indexing operator [] have higher precedence than the dereference operator *. — Some programmer dude
– Some programmer dude, Commented Oct 20, 2015 at 5:35
You should really, really learn the basics of the language. This is trivially answerable using an operator precedence table… — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Oct 20, 2015 at 5:53

Constantin · Accepted Answer · 2015-10-20 05:33:55Z

1

Compare these two lines:

printf("%d %d %d\n", (*p)[0], (*p)[1], (*p)[2]);  //7, 6, 9
printf("%d %d\n", *q[0], *q[1]);  // 4, 7

In the first line you have dereferenced the pointer first and then accessing the index - in your second line you're missing the parenthesis. Changing it to:

  printf("%d %d %d\n", (*p)[0], (*p)[1], (*p)[2]);  // 7, 6, 9
  printf("%d %d %d\n", (*q)[0], (*q)[1]);  // 4, 5

Will output the values as you expect.

answered Oct 20, 2015 at 5:33

Constantin

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Comments

RajSharma · Accepted Answer · 2015-10-20 05:33:40Z

0

it is pointing to the second array in x.

If you want to get all elements of first row you can use something like this.

#include <stdio.h>
int main(void)
{
int x[2][3] =
{
{4, 5, 2},
{7, 6, 9}
};
int (*p)[3] = &x[1];
int (*q)[3] = x;
printf("%d %d %d\n", (*p)[0], (*p)[1], (*p)[2]);  //4, 5, 2
printf("%d %d %d\n", *q[0], q[0][1],q[0][2]);  // 4, 5,2
return 0;
}

answered Oct 20, 2015 at 5:33

RajSharma

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Comments

ameyCU · Accepted Answer · 2015-10-20 05:34:04Z

0

Output you get is valid .

*q[1] is equivalent to q[1][0] . And therefore , value at that index is 7 which you get as output .

To get 5 you can write like this q[0][1] or the other way to represent (*q)[1] .

answered Oct 20, 2015 at 5:34

ameyCU

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Comments

Lundin · Accepted Answer · 2015-10-20 06:50:47Z

0

It's an operator precedence issue. [] has higher priority than unary *. If [] is applied to an array pointer, it will perform pointer arithmetic by the same rules as for regular pointers.

For any pointer type, ptr[i] equals *(ptr + i), and ptr+i means "give me the address of ptr plus i*sizeof(*ptr) bytes.

So in your case q[1] means "give me the address of q + 1*sizeof(*q), where the contents of q is an array of 3 integers. So you end up pointing at the beginning of the 2nd array.

answered Oct 20, 2015 at 6:50

Lundin

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