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:
However, on zooming in, the difference is observable:
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.