Two Dimensional Array Implementation Using Double Pointer

Question

Please consider the following code:

#include 
#include 

#define NUM_ARRAYS     4
#define NUM_ELEMENTS   4
#define INVALID_VAL         


        

Answer 1

*(ptr+i) is equals to ptr[i] and *(ptr+1) is ptr[1].

You can think, a 2-D array as array of array.

• ptr points to complete 2-D array, so ptr+1 points to next 2-D array.

---

In figure below ptr is 2-D and number of columns are 3

Original figure made by Mr. Kerrek SB, here , you should also check!

+===============================+==============================+====
|+---------+----------+--------+|+----------+---------+--------+|
||ptr[0,0] | ptr[0,1] | ptr[0,2]|||ptr[1,0] |ptr[1,1] | ptr[1,2]|| ...
|+---------+----------+--------+++----------+---------+--------++ ...
|            ptr[0]             |           ptr[1]              |
+===============================+===============================+====
   ptr

*(*ptr+1) = *( ptr[0] + 1 ) = ptr[0][1]

Understand following:

ptr points to complete 2-D.

*ptr = *(ptr + 0) = ptr[0] that is first row.

*ptr + 1 = ptr[1] means second row

*(*ptr+1) = *(*(ptr + 0) + 1 ) = *(ptr[0] + 1) = ptr[0][1]

Array 0 Elements:
1  2  3  4 

And GDB Output:

(gdb) p *(*ptr+1)
$1 = 2  

that is correct 2 this can be read using ptr[0][1].

Answer 2

Simplest way for creating 2-dimensinal array using pointer,assigning values and accessing elements from the array.

#include<stdio.h>
#include<stdlib.h>

int main()
{
int i,j;
int row,col;
printf("Enter the values for row and col:\n");
scanf("%d%d",&row,&col);
int **arr=(int**)malloc(row*(sizeof(int*)));
for(i=0;i<row;i++)
{
    *(arr+i)=(int*)malloc(sizeof(int)*col);
            //You can use this also. Meaning of both is same.
            //arr[i]=(int*)malloc(sizeof(int)*col);
}
for(i=0;i<row;i++)
for(j=0;j<col;j++)
{
    arr[i][j]=0;
}
for(i=0;i<row;i++)
{
    for(j=0;j<col;j++)
    {
        printf("%d ",arr[i][j]);
    }
    printf("\n");
}
}

Answer 3

*ptr + 1 is address of ptr[0][1]

Because, we know

ptr[0][1] == *(*(ptr+0)+1)

Now putting ampersand on both side

&ptr[0][1] == &*(*(ptr+0)+1)

You know that ampersand with asterisk is like plus sign with minus sign? They get cancelled. So this becomes

&ptr[0][1] == (*(ptr+0)+1)

Which is same as

&ptr[0][1] == *(ptr+0)+1

Which is again same as

&ptr[0][1] == *ptr+1

This is my first statement.

Answer 4

Unless you mistypes, (*ptr + 1) is equivalent to *(ptr + 0) + 1 which is a pointer to the second element in the first block.

Answer 5

                                 (*ptr)      (*ptr+1)     (*ptr+2)
                                   |            |            |
             __________      ______v____________v____________v____________
  ptr------>|   *ptr   |--->|  *(*ptr)   |  *(*ptr+1)  |*(*ptr+2) |       |
            |__________|    |____________|_____________|__________|_______|
 (ptr+1)--->| *(ptr+1) |     ____________ _____________ __________________
            |__________|--->|*(*(ptr+1)) |*(*(ptr+1)+1)|          |       |
            |          |    |____________|_____________|__________|_______|
            |__________|          ^             ^
                                  |             |
                              *(ptr+1)     *(ptr+1)+1

2D array with double pointers that means that you have a main array and the elements of the main array are pointers (or addresses) to a sub arrays. As indicated in above figure

so if you have defined a double pointer as a pointer of this 2D array let's say int **ptr

so ptr is ponting to the main array which will contains pointers to sub arrays. ptr is ponting to the main array that's means ptr is pointing to the first element of the main array so ptr + 1 is pointing to the second element of the main array.

*ptr this means the content of the first element which the ptr is pointing on. And it is a pointer to a subarray. so *ptr is a pointer to the first subarray (the subarray is an array of int). so *ptr is pointing to the first element in the first subarray. so *ptr + 1 is a pointer to the second element in the first subarray