Here we are discussing pointers and 2D array. While we are discussing the 2D array, the representation of the 2D array is in matrix form. Example:
int arr[3][4] = {
{11,22,33},
{55,66,77},
{11,88,99}
};The 2D array can be visualized in matrix form:
COL 0 COL 1 Col 2
ROW 0 11 22 33
ROW 1 55 66 77
ROW 2 11 88 99We use terminology like rows and columns when talking about arrays. Because computer memory is linear and does not have rows and columns, this concept is just theoretical. So, how are 2-D arrays truly kept in memory? Arrays are stored in C in row-major order. This simply means that row 0 is saved first, followed by row 1, row 2, and so on.
The following representation shows a 2D array in memory:
0-th 1-st 2-nd
1D Array 1D Array 1D Array
==============|==============|=============
11 22 33 55 66 77 11 88 99
100 104 108 112 114 116 120 124 128Here is the most important concept you need to remember in the multi-dimensional array:
A 2-D array is actually a 1-D array in which each element is itself a 1-D array. So arr is an array of 3 elements where each element is a 1-D array of 3 integers.
Important
In this case, of a 2D array, the 0th element is a 1D array. As a result, the type or base type of arr in the above example is a pointer to an array of four integers. Because pointer arithmetic is done in terms of the pointer’s base size. If arr points to address 2000, then arr + 1 points to address 2016 (i.e. 2000 + 3*4) in the case of arr.
We know that the array’s name is a constant pointer pointing to the array’s 0th element. The 0th element in a two-dimensional array is a one-dimensional array. In the case of a 2-D array, the array name is a pointer to the 0th 1-D array. As a result, arr is a pointer to a four-element array in this situation. If the address of the 0th 1-D is 2000, then (arr + 1) represents the address 2012, and (arr + 2) represents the address 2028, according to pointer arithmetic.
From the above discussion, we can conclude that:
arr points to 0th 1-D array.(arr + 1) points to 1st 1-D array.(arr + 2) points to the 2nd 1-D array.
arr => 11 22 33
arr + 1 => 55 66 77
arr + 2 => 11 88 99We can write the following in general:
(arr + i) is the index of the ith 1-D array.
Dereferencing a pointer to an array returns the array’s base address, as we discussed earlier in this chapter. Dereferencing arr yields arr, whose base type is (int). Similarly, when we decode (arr+i) we get *(arr+i). In general, we can state the following:
The base address of the ith 1-D array is pointed to by *(arr+i).
It’s worth noting that while type (arr + i) and *(arr+i) both refer to the same address, their basic types are completely different. (arr + i) is a pointer to a three-int array, whereas *(arr + i) is a pointer to int or (int*).
So, how can you access individual items in a 2-D array using arr?
We should be able to access elements of ith 1-D array using pointer arithmetic because *(arr + i) points to the base address of every ith 1-D array and is of base type pointer to int.
Let’s have a look at how we can accomplish this:
The address of the
0th element of the 1-D array is pointed to by *(arr + i). As a result, *(arr +i) + 1 denotes the address of the 1-D array’s first element. *(arr + i) + 2 is the address of the 1-D array’s second element.
As a result, we can deduce that:
*(arr + i) + j points to the base address of the ith 1-D array’s jth element.
We may get the value of the jth element of the ith 1-D array by dereferencing *(arr + i) + j.
*( *(arr + i) + j)
We can obtain the value of the jth element of the ith 1-D array using this expression.
Furthermore, the subscript notation is equal to the pointer notation *(*(arr + i )+ j).
The following program shows how to use pointer notation to retrieve the values and addresses of elements in a 2-D array.
#include<stdio.h>
int main()
{
int arr[3][4] = {
{11,22,33},
{55,66,77},
{11,88,99}
};
int i, j;
for(i = 0; i < 3; i++)
{
printf("Address of %d th array %u \n",i , *(arr + i));
for(j = 0; j < 4; j++)
{
printf("arr[%d][%d]=%d\n", i, j, *( *(arr + i) + j) );
}
printf("\n\n");
}
// signal to operating system program ran fine
return 0;
}Output:
Address of 0 th array 2686736
arr[0][0]=11
arr[0][1]=22
arr[0][2]=33
Address of 1 th array 2686752
arr[1][0]=55
arr[1][1]=66
arr[1][2]=77
Address of 2 th array 2686768
arr[2][0]=11
arr[2][1]=88
arr[2][2]=99
Assigning 2-D Array to a Pointer Variable
You can assign the array’s name to a pointer variable, however, unlike with a 1-D array, you’ll require a pointer to an array rather than a pointer to int or (int *). Here’s an illustration:
int arr[2][3] = {
{33, 44, 55},
{11, 99, 66}
};Always keep in mind that a 2-D array is a 1-D array with each element being a 1-D array. As a result, arr is a two-element array with each element being a one-dimensional arr of three integers. As a result, you’ll need a pointer to an array of 3 integers to store arr’s base address.
Similarly, if a 2-D array includes three rows and four columns, such as int arr[3][4], you’ll need a pointer to a four-int array.
int (*p)[3];Here, p is a pointer to a three-int array. So, according to pointer arithmetic, p+i corresponds to the ith 1-D array, whereas p+0 corresponds to the 0th 1-D array, p+1 to the 1st 1-D array, and so on. (p+i) has a pointer to a three-int array as its base type. We can access the base address of ith 1-D array if we dereference (p+i), but now the base type of *(p + i) is a pointer to int or (int *). To get the address of the jth element of the ith 1-D array, simply add j to *(p +i) as a result, *(p +i) + j denotes the address of the jth element of the ith 1-D array.
The following program demonstrates how to access elements of a 2-D array using a pointer to an array.
#include<stdio.h>
int main()
{
int arr[3][3] = {
{11,22,33},
{55,66,77},
{11,88,99}
};
int i, j;
int (*p)[3];
p = arr;
for(i = 0; i < 3; i++)
{
printf("Address of %d th array %u \n",i , p + i);
for(j = 0; j < 4; j++)
{
printf("arr[%d][%d]=%d\n", i, j, *( *(p + i) + j) );
}
printf("\n\n");
}
// signal to operating system program ran fine
return 0;
}Expected Output:
Address of 0 th array 2686736
arr[0][0]=11
arr[0][1]=22
arr[0][2]=33
Address of 1 th array 2686752
arr[1][0]=55
arr[1][1]=66
arr[1][2]=77
Address of 2 th array 2686768
arr[2][0]=11
arr[2][1]=88
arr[2][2]=99
