Question

I have an image compression application that now has two different versions of memory allocation systems. In the original one, malloc is used everywhere, and in the second one, I implemented a simple pool-allocator, that just allocates chunk of memory and returns parts of that memory to myalloc() calls.

We’ve been noticing a huge memory overhead when malloc is used: At the height of its memory usage, the malloc() code requires about 170 megabytes of memory for a 1920x1080x16bpp image, while the pool allocator allocates just 48 megabytes, of which 47 are used by the program.

In terms of memory allocation patterns, the program allocates a lot of 8byte(most), 32-byte(many) and 1080byte-blocks(some) with the test image. Apart from these, there are no dynamic memory allocations in the code.

The OS of the testing system is Windows 7 (64 Bit).

How did we test memory usage?

With the custom allocator, we could see how much memory is used because all malloc calls are defered to the allocator. With malloc(), in Debug mode we just stepped through the code and watched the memory usage in the task manager. In release mode we did the same, but less fine grained because the compiler optimizes a lot of stuff away so we couldn’t step through the code piece by piece (the memory difference between release and debug was about 20MB, which I would attribute to optimization and lack of debug information in release mode).

Could malloc alone be the cause of such a huge overhead? If so, what exactly causes this overhead inside malloc?

Answer 1

First at all malloc aligns the pointers to 16 byte boundaries. Furthermore they store at least one pointer (or allocated length) in the addresses preceding the returned value. Then they probably add a magic value or release counter to indicate that the linked list is not broken or that the memory block has not been released twice (free ASSERTS for double frees).

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

int main(int ac, char**av)
{
  int *foo = malloc(4);
  int *bar = malloc(4);
  printf("%d\n", (int)bar - (int)foo);
}

Return: 32