Using 2D arrays as multiple 1D arrays when using pointers

Question

I used the below program to access 2D arrays using pointers.

    #include<stdio.h>

void main()
{

  int num[3][2]={ {00,01},{10,11},{20,21} }; 
  int i,j;
  printf("-----------------------------------");
  /* Treating 2d array as 1d array of each row */
  for(i=0;i<3;i++)
  {
        printf("\nThe Address(&num[%d]) is %u \n \n<= WHICH IS THE SAME AS =>\n \n(num[%d]) %u \t",i,&num[i],i,num[i]);
        printf("[*(num+%d)] %u \t [(num+%d)] %u ",i,*(num+i),i,(num+i)); 
        printf("\n The Value is %d \n",*num[i]);
        printf("\n -----------------------------------");
  }
}

And this is the Output:

    -----------------------------------
The Address(&num[0]) is 2140353424 

<= WHICH IS THE SAME AS =>

(num[0]) 2140353424     [*(num+0)] 2140353424    [(num+0)] 2140353424 
 The Value is 0 

 -----------------------------------
The Address(&num[1]) is 2140353432 

<= WHICH IS THE SAME AS =>

(num[1]) 2140353432     [*(num+1)] 2140353432    [(num+1)] 2140353432 
 The Value is 10 

 -----------------------------------
The Address(&num[2]) is 2140353440 

<= WHICH IS THE SAME AS =>

(num[2]) 2140353440     [*(num+2)] 2140353440    [(num+2)] 2140353440 
 The Value is 20 

 -----------------------------------

I understood what happens in the program and I know that *(num+i) is used to access each row's address.

But, why both *(num+i) and (num+i) point to the same address?

To access the value at that particular row (i'th row) we use **(num+i) which makes sense because *(num+i) points to the 1st 1D array's row's address and we can use another indirection operator to dereference that pointer.

But how come both *(num+i) and (num+i) point to the same address? Is this compiler dependent? Or some undefined behavior?

Please provide as much information as possible.

where is the problem? it is ok, *(num + i) = (num + i)[0] and *((num + i)[0]) = num[i][0] — elhadi dp ıpɐɥןǝ
– elhadi dp ıpɐɥןǝ, Commented Sep 15, 2014 at 13:57

macfij · Accepted Answer · 2014-09-16 16:26:37Z

2

Multi-dimensional arrays are stored contiguously in memory. Here's a memory layout of your array:

 [0][0] [0][1] [1][0] [1][1] [2][0] [2][1]
+------+------+------+------+------+------+
|  00  |  01  |  10  |  11  |  20  |  21  |
+------+------+------+------+------+------+
|_____________|_____________|_____________|
^             ^             ^
num[0]        num[1]        num[2]

On my machine int has size of 4 bytes.
num has type int (*)[2] - it is a pointer to array of 2 ints. Statement num+1 is a pointer arithmetic and results in adding 8 bytes (that's the size of two ints) to the address where num array begins.
After dereference - *(num+1) it is still a pointer and it points to the same address as num+1. You can check this short code to see what is happening:

#include <stdio.h>

int main(void)
{
    int num[3][2]={ {00,01},{10,11},{20,21} }; 
    int (*fp)[2] = num+1;
    int* fpp = *(num+1);
    printf("%d\n", (void*)fpp == (void*)fp);
    return 0;
}

The output is 1 - those pointers hold the same memory address.

edited Sep 16, 2014 at 16:26

answered Sep 15, 2014 at 14:48

macfij

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1 Comment

Abinesh S Prabhakaran Over a year ago

Thanks that code diagram helped me understand your post.

python · Accepted Answer · 2014-09-15 14:56:33Z

0

But, why both *(num+i) and (num+i) point to the same address?

It's because:

The array name is a pointer (address) to the first element of the array.
array[x] equals to *(array+x)

answered Sep 15, 2014 at 14:56

python

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