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what happens when you don’t free memory after using malloc()

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Pre-requisite: Dynamic memory allocation in C
The “malloc” or “memory allocation” method is used to dynamically allocate a single large block of memory with the specified size. It returns a pointer of type void which can be cast into a pointer of any form. It initializes each block with a default garbage value.

ptr = (cast-type*) malloc(byte-size)

For Example: 

ptr = (int*) malloc(100 * sizeof(int));
Since the size of int is 4 bytes, this statement will allocate 400 bytes of memory. And, the pointer ptr holds the address of the first byte in the allocated memory. 

But the memory allocation using malloc() is not de-allocated on its own. So, “free()” method is used to de-allocate the memory. But the free() method is not compulsory to use. If free() is not used in a program the memory allocated using malloc() will be de-allocated after completion of the execution of the program (included program execution time is relatively small and the program ends normally). Still, there are some important reasons to free() after using malloc()

  • Use of free after malloc is a good practice of programming.
  • There are some programs like a digital clock, a listener that runs in the background for a long time and there are also such programs which allocate memory periodically. In these cases, even small chunks of storage add up and create a problem. Thus our usable space decreases. This is also called “memory leak”. It may also happen that our system goes out of memory if the de-allocation of memory does not take place at the right time.

So, the use of the free() functions depends on the type of the program but it is recommended to avoid the unwanted memory issues like a memory leak.
Below is the program to illustrate the use of free() and malloc():


// C program to illustrate free()
// and malloc() function
#include <stdio.h>
#include <stdlib.h>
// Driver Code
int main()
    // Pointer to hold base address
    // of memory block
    int* p;
    int n, i;
    // Number of element
    n = 3;
    // Dynamically allocate memory
    // using malloc()
    p = (int*)malloc(n * sizeof(int));
    // Check if memory allocation is
    // successful or not.
    if (p == NULL) {
        printf("Memory allocation"
               " failed.\n");
    else {
        // Memory allocation successful
        printf("Memory allocation "
               "successful using"
               " malloc.\n");
        // Insert element in array
        for (i = 0; i < n; i++) {
            p[i] = i + 2;
        // Print the array element
        printf("The array elements"
               " are:\n");
        for (i = 0; i < n; i++) {
            printf("%d ", p[i]);
        // De-allocate the allocated
        // memory using free()
    return 0;


Memory allocation successful using malloc.
The array elements are:
2 3 4 

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Last Updated : 05 Oct, 2021
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