malloc using 4 bytes for char - memory

Question

I am writing a code to examine how memory is managed between stack and heap. for a course work.
#include<stdio.h>
#include<stdlib.h>
#define NUM_OF_CHARS 100
// function prototype
void f(void);
int main()
{
f();
return 0;
}
void f(void)
{
char *ptr1;
ptr1 = (char *) malloc(NUM_OF_CHARS * sizeof(int));
printf("Address array 1: %016lx\n", (long)ptr1);
char *ptr2;
ptr2 = (char *) malloc(NUM_OF_CHARS * sizeof(int));
printf("Address array 2: %016lx\n", (long)ptr2);
}
when I run this code I get the following:
Address array 1: 000000000209e010
Address array 2: 000000000209e1b0
my expectation was to see a difference in the address of 100 bytes, but the difference is 416 bytes, when I changed the NUM_OF_CHARS to any other value (200,300,...) the result was always (NUM_OF_CHARS*4 + 16), so it seams like malloc is allocating 4 bytes for each char rather one byte plus 16 bytes of some overhead.
can anyone explain what is happening here?

Answer 1

Memory allocation is platform/compiler dependent. The only thing malloc ensures is that it allocates enough memory for what you are asking and nothing more.
There is no guarantee that your addresses will be contiguous due to memory alignment
Also, you are allocating by size of ints and not char in your code. This is most likely the reason why you see a NUM_OF_CHARS*4 difference, while the remaining difference can be attributed to padding.