Write C code to dynamically allocate one, two and three dimensional arrays (using malloc())

Write C code to dynamically allocate one, two and three dimensional arrays (using malloc())

Its pretty simple to do this in the C language if you know how to use C pointers. Here are some example C code snipptes….

One dimensional array

 

#include <stdio.h>
#include <stdlib.h>
 
int main()
{
	   int i, count = 0, no_of_elements=5;
	 
	   // Allocating dynamic array	 
	   int *array = malloc(no_of_elements * sizeof(int));
	    
	   // Accessing one dimensional array  
	   for (i = 0; i<no_of_elements; i++)
	         array[i] = ++count;  // OR *(arr+i) = ++count
	 
	   for (i = 0; i <  no_of_elements; i++)
	         printf("%d ", array[i]);
	
	   return 0;
}

 

 

Two dimensional array

There are more than one methods to allocate two dimensional array dynamically but we use only pointer method. 

 

#include <stdio.h>
#include <stdlib.h>
 
#define ROW       3
#define COLUMN    4   
 
int main()
{
    int i, j, count=0;
 
 	// Allocating two dimensional dynamic array	 
    int **arr = (int **)malloc(ROW * sizeof(int *));
    for (i=0; i<ROW; i++)
         arr[i] = (int *)malloc(COLUMN * sizeof(int));
 
    // Accessing two dimensional array
    for (i = 0; i <  ROW; i++)
    {
    	for (j = 0; j < COLUMN; j++)
    	{
    		 arr[i][j] = ++count;  // OR *(*(arr+i)+j) = ++count
    		 printf("%d ", arr[i][j]);	
    	}  
    }
 
    return 0;
}

 

 

Three dimensional array

 

#include<stdio.h>
#include<malloc.h>
 
#define AXIS_X 3
#define AXIS_Y 4
#define AXIS_Z 5
 
int main()
{
	int ***p,i,j,k,count=0;
	
	// Allocating three dimensional dynamic array
	p=(int ***) malloc(AXIS_X * sizeof(int ***));
	
	for(i=0;  i<AXIS_X;  i++)
	{
		p[i]=(int **)malloc(AXIS_Y * sizeof(int *));
		
		for(j=0;  j<AXIS_Y;  j++)
		{
			p[i][j]=(int *)malloc(AXIS_Z * sizeof(int));
		}
	}
	
	// Accessing three dimensional array
	for(k=0;  k<AXIS_Z;  k++)
	{
		for(i=0;  i<AXIS_X;  i++)
		{
			for(j=0;  j<AXIS_Y;  j++)
			{
				*(*(*(p+i)+j)+k)= ++count;	// OR *(*(*(p+i)+j)+k) = ++count
				printf("%d ",p[i][j][k]);	
			}
		}
	}
	
	
	return 0;
}