Pointer to Arrays

Question

i am using a pointer to an array of 4 integers as int(*ptr)[4]

using the following code in which m pointing to a 2-D array using this pointer

int arr[3][4]={{1,2,3,4},{5,6,7,8},{9,10,11,12}};
int (*ptr)[4]= arr[4];
int m= (*ptr)[2];

what will be the value in "m"...

i need to find the value of element arr[1][2] how can i get it using pointer ptr?

• Carl Norum: Format your code next time. Is this a homework question? Tag it that way if it is, and tell us what you've tried and what your exact problem is.
• Carl Norum: Also, what's stopping you from running it and trying (it won't work, by the way).
• Wix: On the second line 'arr[4]' isn't that an index out of range? The max size of arr has been declared as 3. So the maximum index should be 2 as it is zero-based.
• user363801: make ir arr[2] even then its not giving the correct output...
• user363801: ptr= arr[2] gives an error.. why???

Answer 1

Multi-dimensional arrays are really one dimensional arrays with a little syntaxic sugar. The initialization you had for ptr wasn't an address. It should have been

int *ptr[4] = { &arr[0][0], &arr[1][0], &arr[2][0], &arr[3][0]};

You can also leave the 4 out and just use

int *ptr[] = { &arr[0][0], &arr[1][0], &arr[2][0], &arr[3][0]};

I made a few modifications to your code below. Note the two sections with the printf. They should help to demonstrative how the values are actually laid out in memory.

#define MAJOR 3
#define MINOR 4
int arr[MAJOR][MINOR]={{1,2,3,4},{5,6,7,8},{9,10,11,12}};
int (*ptr)[4];
int *p = &arr[0][0];
// init ptr to point to starting location in arr
for(i = 0; i < MAJOR; i++) {
 ptr[i] = &arr[i][0];
}
// print out all the values of arr using a single int *
for(i = 0; i < MAJOR * MINOR; i++) {
 printf(" %d", *(p + i) );
}
for(i = 0; i < MAJOR; i++) {
 for(j = 0; j < MINOR; j++) {
 printf( " %d", *(p + i * MAJOR + j) );
 }
 printf("\n");
}

Answer 2

Ptr doesn't point to anything and probably doesn't even compile, since [4] is outside the bounds of either the [3] or the [4] of the original array.

If you need to find arr[1][2] then you need int (*ptr)[4] = arr[1]; int m = ptr[2];

• AnT stands with Russia: int (*ptr)[4] = &arr[1]; int m = (*ptr)[2]; Note the extra & and *. Your variant will not even compile.
• user363801: hey anDrey tell me one thing int (*ptr)[4]= &arr[4] will be out of bound?
• Carl Norum: @nitinpuri, yes it will be out of bounds - arr only has 3 elements, so the maximum index is 2.
• Wix: arr was declared with the size [3] and [4]. With only one index it refers to the first size [3]. Since arrays in C/C++ are zero based the maximum value allowed there would be 2. And to clarify what DeadMG refers to, you need to use the '&' to get the reference of arr at the specified index.
• user363801: first part is done but int m= (*ptr)[2] does not point to element 3 of 4 element array..

Answer 3

The are multiple ways of accessing the value of element arr[1][2].

The method used in the code below is called pointer to pointer. By storing the address of the first element of pa[] in pb, one can use pb to access elements of aa[][]. One can use array notation aa[1][2], or one can use pointer notation *((*pb+1)+2). Both the notation approaches yield equivalent results.

#include<stdio.h>
int main()
{
 int aa[3][4]={{1,2,3,4},{5,6,7,8},{9,10,11,12}};
 int *pa[3];
 for(int i=0;i<3;i++)
 pa[i]=aa[i];
 pb=pa;
 printf("pb[1][2]=%d\n",pb[1][2]); /*Array Notation*/
 printf("arr[1][2]=%d",*(*(pb+1)+2)); /*Pointer Notation*/
}