Written: August 22, 2026
Dynamic allocation lets the size come from input instead of a fixed compile-time array length.
This walkthrough keeps the logic small and readable so you can type it, run it, and then change one input at a time to see what happens.
What you will build
You will allocate n ints, fill them, print them, then free the block.
- Read n
- p = malloc(n * sizeof *p) or calloc(n, sizeof *p)
- Check for NULL
- Use the block, then free(p)
Working C example
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int n, i;
int *p;
printf("n: ");
scanf("%d", &n);
p = malloc((size_t)n * sizeof *p);
if (!p) {
printf("Allocation failed\n");
return 1;
}
for (i = 0; i < n; i++) p[i] = i + 1;
for (i = 0; i < n; i++) printf("%d ", p[i]);
printf("\n");
free(p);
return 0;
}
How the logic works
Read the program top to bottom: includes and main first, then the statements that change variables, then the print that proves the result.
If your output looks wrong, print the variables before and after the critical lines. That single habit catches most beginner bugs faster than rewriting the whole file.
calloc zeroes memory; malloc does not. Always free exactly once for each successful allocation path.
Common mistakes
Forgetting a semicolon, using the wrong format specifier in printf/scanf, and mixing up assignment (=) with comparison (==) are the usual culprits.
Compile with warnings enabled (`gcc -Wall`) so the compiler points at risky casts and unused variables before you chase them by hand.
Try this next
Change the sample inputs, add a second test case, and briefly note what stayed the same. Teaching yourself with tiny experiments sticks better than copying a longer program you never run.
When you can explain every line without looking, you are ready for the next exercise in the series.
Keep learning
If this walkthrough on malloc calloc free in c helped, open the code again and change one input or assumption. Small experiments beat rereading the same example.
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