[9fans] Here is a silly question..

[9fans] Here is a silly question..

[ISHWAR RATTAN]

How does one do a time-out using alarm() call? It seems
that syscall read() is not aborted by alarm()..

   int rd, try = 0;
   char buf[1024];

   while(try < 2) {
      alarm(2 * 1000);
      rd = read(0, buf, 1024);
      alarm(0);
      if(rd > 0) {
 	write(1, buf, rd);
 	break;
      }
   }

-ishwar

Re: [9fans] Here is a silly question..

[Russ Cox]

> How does one do a time-out using alarm() call? It seems
> that syscall read() is not aborted by alarm()..

Yes it is.

Russ

Re: [9fans] Here is a silly question..

[erik quanstrom]

you need to increment try each time through the loop.
this works for me:
#include <u.h>
#include <libc.h>

void
fu(void)
{
	int rd, try;
	char buf[1024];

	for(try=0; try < 2; try++){
		alarm(2 * 1000);
		rd = read(0, buf, 1024);
		alarm(0);
		if(rd > 0) {
			write(1, buf, rd);
			break;
		}
	}
}

void
main(void)
{
	fu();
	exits("");
}