Dynamic Memory Allocation
Allocation of memory at the time of execution (run time) is known as dynamic memory allocation.
The functions calloc() and malloc() support allocating of dynamic memory.
Dynamic allocation of memory space is done by using these functions when value is returned by functions and assigned to pointer variables.
In this case, variables get allocated only if your program unit gets active.
It uses the data structure called heap for implementing dynamic allocation.
There is memory reusability and memory can be freed when not required.
It is more efficient.
In this memory allocation scheme, execution is slower than static memory allocation.
Here memory can be released at any time during the program.
Example
Following program computes the sum of even numbers and odd numbers in a set of elements using dynamic memory allocation functions −
#include<stdio.h>
#include<stdlib.h>
void main(){
//Declaring variables, pointers//
int i,n;
int *p;
int even=0,odd=0;
//Declaring base address p using malloc//
p=(int *)malloc(n*sizeof(int));
//Reading number of elements//
printf("Enter the number of elements : ");
scanf("%d",&n);
/*Printing O/p -
We have to use if statement because we have to check if memory
has been successfully allocated/reserved or not*/
if (p==NULL){
printf("Memory not available");
exit(0);
}
//Storing elements into location using for loop//
printf("The elements are : \n");
for(i=0;i<n;i++){
scanf("%d",p+i);
}
for(i=0;i<n;i++){
if(*(p+i)%2==0){
even=even+*(p+i);
}
else{
odd=odd+*(p+i);
}
}
printf("The sum of even numbers is : %d\n",even);
printf("The sum of odd numbers is : %d\n",odd);
}Output
Enter the number of elements : 4 The elements are : 35 24 46 12 The sum of even numbers is : 82 The sum of odd numbers is : 35