🚀 go-pugleaf

In <vkohklaizj.fsf@atari.connectinc.com> bji@connectinc.com (Bryan Ischo) writes:

>In article <danpop.839504721@news.cern.ch> Dan.Pop@cern.ch (Dan Pop) writes:
>   In <4ubnat$75u@freeside.fc.net> mrbig@freeside.fc.net (Mike Big) writes:
>
>   >int *flag_pt;
>   >
>   >int main()
>   >{
>   >  struct itimerval timer_length;
>   >  int  flag;
>   >  
>   >  flag_pt=&flag;
>   >  /* specify signal handler */
>   >  signal(SIGALRM, fca);
>   >
>   >  /* set the signal to repeat every 3 seconds */
>   >  timer_length.it_value.tv_sec=3;     /* initial time */
>   >  timer_length.it_value.tv_usec=0;
>   >  timer_length.it_interval.tv_sec=3;  /* repeat */
>   >  timer_length.it_interval.tv_usec=0;
>   >  setitimer(ITIMER_REAL, &timer_length, NULL);
>   >  
>   >  flag=time(NULL); /* this is so that the compiler won't substitute a 
>   >			  constant for flag as it might if I did something
>   >                      like flag=1; */
>   >
>   >  /* count until interrupted (after 3 seconds) */
>   >  counter=0;
>   >  while(flag){ /* does NOT exit when using gcc -O2 */
>
>[more skippage]
>
>   >    counter++;
>   >    /* uncomenting the printf below will fix things. weird heh */
>
>   There is nothing weird here.  flag is accessible via a global pointer 
>   so printf might change its value, as a side effect.  The compiler can't
>   tell whether this will happen or not, so it cannot assume that flag is
>   a loop invariant in the presence of a function call inside the loop.
>   The above-mentioned optimization is illegal now.  A smarter compiler
>   might have more information about printf and decide from its invocation
>   that it cannot modify flag and still do the optimization :-)
>
>   >    /* printf("counter=%ld flag=%d\n",counter,flag); */ 
>   >  }
>   >
>   >  exit(0);
>
>[skipped]
>
>Just out of curiosity ... why can't the compiler be sure that 'flag' won't be
>modified by printf? 

Because flag is accessible via the flag_pt pointer, which is a global
variable, hence available to any function called by the program.
 
>Shouldn't the compiler be smart enough to know that flag can't be modified
>even IF the printf is in the loop, given how smart it's been about everything
>else?

On the contrary, it is smart enough to know that flag CAN be modified by
a function called within the loop.

Dan
--
Dan Pop
CERN, CN Division
Email: Dan.Pop@cern.ch 
Mail:  CERN - PPE, Bat. 31 R-004, CH-1211 Geneve 23, Switzerland