malloc using 4 bytes for char

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?

Copyright Notice：Content Author:「user3261224」，Reproduced under the CC 4.0 BY-SA copyright license with a link to the original source and this disclaimer.
Link to original article：https://stackoverflow.com/questions/28734226/malloc-using-4-bytes-for-char