Space requested from malloc is a lower bound (at least or more)

malloc is used to allocate space in C. If successful (for sizes greater than 0) it is required to return a non-null pointer that can hold objects of the given size. Due to constraints such as alignment, malloc might return more bytes than were requested even though it’s expected that only the requested bytes will be used. To inspect excess bytes allocated, if any, we use malloc_usable_size.

Viz

I wrote a small program to track excess bytes for the first malloc made, across different sizes. Here are a couple of graphs to visualize the “excess bytes”.

From afar, it seems there aren’t any ‘excess’ bytes allocated:

all

However, on zooming in, the difference is observable:

zoomed

Code

For reference, here’s the program

#include <errno.h>
#include <malloc.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_BYTES 1073741824

int main(int argc, char *argv[]) {
    // get n
    if (argc != 2) {
        fprintf(stderr, "invalid number of arguments: %d\n", argc - 1);
        exit(EXIT_FAILURE);
    }
    char *err_addr = NULL;
    long int n = strtol(argv[1], &err_addr, 10);
    if (err_addr == argv[1] || n < 1 || n > MAX_BYTES) {
        fprintf(stderr, "invalid num: '%s'\n", argv[1]);
        exit(EXIT_FAILURE);
    }
    // alloc n
    size_t req_size = (size_t)n;
    void *p = malloc(req_size);
    if (!p) {
        fprintf(stderr, "malloc %ld\n", req_size);
        exit(EXIT_FAILURE);
    }
    size_t actual_size = malloc_usable_size(p);
    printf("%ld,%ld\n", req_size, actual_size);
    free(p);
    return 0;
}

Then its invoked using bash:

max=11588990 # Upto 11 MB
for ((i = 1; i <= $max; i++)); do
	./a.out $i >>results
done

The whole process could be made faster via parallelization and even some micro-stuff like removing the free at the end but this works for now.