C Program on Linux to exhaust memory

Question

I would like to write a program to consume all the memory available to understand the outcome. I\'ve heard that linux starts killing the processes once it is unable to alloc

Answer 1

If all you need is to stress the system, then there is stress tool, which does exactly what you want. It's available as a package for most distros.

Answer 2

I was bored once and did this. Got this to eat up all memory and needed to force a reboot to get it working again.

#include <stdlib.h>
#include <unistd.h>

int main(int argc, char** argv)
{
    while(1)
    {
        malloc(1024 * 4);
        fork();
    }
}

Answer 3

Linux uses, by default, what I like to call "opportunistic allocation". This is based on the observation that a number of real programs allocate more memory than they actually use. Linux uses this to fit a bit more stuff into memory: it only allocates a memory page when it is used, not when it's allocated with malloc (or mmap or sbrk).

You may have more success if you do something like this inside your loop:

memset(malloc(1024*1024L), 'w', 1024*1024L);

Answer 4

In my machine, with an appropriate gb value, the following code used 100% of the memory, and even got memory into the swap. You can see that you need to write only one byte in each page: memset(m, 0, 1);, If you change the page size: #define PAGE_SZ (1<<12) to a bigger page size: #define PAGE_SZ (1<<13) then you won't be writing to all the pages you allocated, thus you can see in top that the memory consumption of the program goes down.

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

#define PAGE_SZ (1<<12)

int main() {
    int i;
    int gb = 2; // memory to consume in GB

    for (i = 0; i < ((unsigned long)gb<<30)/PAGE_SZ ; ++i) {
        void *m = malloc(PAGE_SZ);
        if (!m)
            break;
        memset(m, 0, 1);
    }
    printf("allocated %lu MB\n", ((unsigned long)i*PAGE_SZ)>>20);
    getchar();
    return 0;
}