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Memory Allocation Assignment Help

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A program that Demonstrates Dynamic Memory Allocation in C

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#include <stdio.h> #include <stdlib.h> int main() { int *ptr; int n, i, sum = 0; // Input the number of elements printf("Enter number of elements: "); scanf("%d", &n); // Allocate memory dynamically ptr = (int *) malloc(n * sizeof(int)); // Check if memory allocation is successful if (ptr == NULL) { printf("Memory allocation failed. Exiting..."); return 1; } // Input elements and compute their sum printf("Enter %d elements:\n", n); for (i = 0; i < n; i++) { scanf("%d", &ptr[i]); sum += ptr[i]; } // Display the elements and their sum printf("Elements are: "); for (i = 0; i < n; i++) { printf("%d ", ptr[i]); } printf("\nSum of elements = %d\n", sum); // Free dynamically allocated memory free(ptr); return 0; }

In this program:

• We declare a pointer ptr of type int to hold the base address of the dynamically allocated memory.
• We prompt the user to enter the number of elements (n) they want to allocate memory for.
• We use malloc() to dynamically allocate memory for n integers (n * sizeof(int) bytes).
• We check if memory allocation was successful by ensuring ptr is not NULL.
• We then prompt the user to input n integers, compute their sum, and display both the elements and their sum.
• Finally, we free the dynamically allocated memory using free(ptr) to prevent memory leaks.

This example illustrates the basic usage of malloc() for dynamic memory allocation in C. Remember to always check for NULL after calling malloc() to handle cases where memory allocation fails. You can expect such quality programs when you avail our memory allocation in C assignment help.

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