How to get swap size by using system call?

Question

How can I get the size of swap.What system call or library Routines can be used? please help me.

Answer 1

Hi,
Threre's an interface called swap_ctl by wich you can get more information related to virtual memory. Here's an example:
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/stat.h>
#include <sys/swap.h>
#include <sys/param.h>
int
main(int argc, char*argv[])
swaptbl_t *st;
int swap_count, i;
long long swap_total=0, swap_avail=0;
long pgsize_in_kbytes = sysconf(_SC_PAGE_SIZE) / 1024L;
if ((swap_count=swapctl(SC_GETNSWP, NULL)) == -1)
perror("swapctl(SC_GETNSWP)"), exit(-1);
if (swap_count == 0)
(void)printf("No swap files/partitions allocated\n"), exit(0);
* Although it's not particularly clear in the documentation, you're
* responsible for creating a variable length structure (ie. the
* array is within the struct rather than being pointed to
* by the struct). Also, it is necessary for you to allocate space
* for the path strings (see /usr/include/sys/swap.h).
st = (swaptbl_t*)malloc(sizeof(int) + swap_count * sizeof(struct swapent));
if (st == NULL)
perror(argv[0]), exit(-1);
st->swt_n = swap_count;
for (i=0; i < swap_count; i++) {
if ((st->swt_ent.ste_path = (char*)malloc(MAXPATHLEN)) == NULL)
perror(argv[0]), exit(-1);
if ((swap_count=swapctl(SC_LIST, (void*)st)) == -1)
perror("swapctl(SC_LIST)"), exit(-1);
for (i=0; i < swap_count; i++) {
swap_total += st->swt_ent.ste_pages * pgsize_in_kbytes;
swap_avail += st->swt_ent.ste_free * pgsize_in_kbytes;
(void)printf("Total swap = %lld available swap = %lld\n",
swap_total, swap_avail);
return 0;
There's a very interesting document about Solaris Kernel/Performance Stats available at http://www.idiom.com/~gford/admin/howto/perf.html. Very interesting.
Hope this helps