Question

I got some huge files I need to parse, and people have been recommending mmap because this should avoid having to allocate the entire file in-memory.

But looking at ‘top’ it does look like I’m opening the entire file into the memory, so I think I must be doing something wrong. ‘top shows >2.1 gig’

This is a code snippet that shows what I’m doing.

Thanks

#include <stdio.h>
#include <stdlib.h>
#include <err.h>
#include <fcntl.h>
#include <sysexits.h>
#include <unistd.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <sys/mman.h>
#include <cstring>
int main (int argc, char *argv[] ) {
  struct stat sb;
  char *p,*q;
  //open filedescriptor
  int fd = open (argv[1], O_RDONLY);
  //initialize a stat for getting the filesize
  if (fstat (fd, &sb) == -1) {
    perror ("fstat");
    return 1;
  }
  //do the actual mmap, and keep pointer to the first element
  p =(char *) mmap (0, sb.st_size, PROT_READ, MAP_SHARED, fd, 0);
  q=p;
  //something went wrong
  if (p == MAP_FAILED) {
    perror ("mmap");
    return 1;
  }
  //lets just count the number of lines
  size_t numlines=0;
  while(*p++!='\0')
    if(*p=='\n')
      numlines++;
  fprintf(stderr,"numlines:%lu\n",numlines);
  //unmap it
  if (munmap (q, sb.st_size) == -1) {
    perror ("munmap");
    return 1;
  }
  if (close (fd) == -1) {
    perror ("close");
    return 1;
  }
  return 0;
}

Answer 1

No, what you’re doing is mapping the file into memory. This is different to actually reading the file into memory.

Were you to read it in, you would have to transfer the entire contents into memory. By mapping it, you let the operating system handle it. If you attempt to read or write to a location in that memory area, the OS will load the relevant section for you first. It will not load the entire file unless the entire file is needed.

That is where you get your performance gain. If you map the entire file but only change one byte then unmap it, you’ll find that there’s not much disk I/O at all.

Of course, if you touch every byte in the file, then yes, it will all be loaded at some point but not necessarily in physical RAM all at once. But that’s the case even if you load the entire file up front. The OS will swap out parts of your data if there’s not enough physical memory to contain it all, along with that of the other processes in the system.

The main advantages of memory mapping are:

• you defer reading the file sections until they’re needed (and, if they’re never needed, they don’t get loaded). So there’s no big upfront cost as you load the entire file. It amortises the cost of loading.
• The writes are automated, you don’t have to write out every byte. Just close it and the OS will write out the changed sections. I think this also happens when the memory is swapped out as well (in low physical memory situations), since your buffer is simply a window onto the file.

Keep in mind that there is most likely a disconnect between your address space usage and your physical memory usage. You can allocate an address space of 4G (ideally, though there may be OS, BIOS or hardware limitations) in a 32-bit machine with only 1G of RAM. The OS handles the paging to and from disk.

And to answer your further request for clarification:

Just to clarify. So If I need the entire file, mmap will actually load the entire file?

Yes, but it may not be in physical memory all at once. The OS will swap out bits back to the filesystem in order to bring in new bits.

But it will also do that if you’ve read the entire file in manually. The difference between those two situations is as follows.

With the file read into memory manually, the OS will swap parts of your address space (may include the data or may not) out to the swap file. And you will need to manually rewrite the file when your finished with it.

With memory mapping, you have effectively told it to use the original file as an extra swap area for that file/memory only. And, when data is written to that swap area, it affects the actual file immediately. So no having to manually rewrite anything when you’re done and no affecting the normal swap (usually).

It really is just a window to the file: