gettime: fix overflow in cycle to usec conversion
authorJens Axboe <axboe@fb.com>
Wed, 17 Dec 2014 05:40:37 +0000 (22:40 -0700)
committerJens Axboe <axboe@fb.com>
Wed, 17 Dec 2014 05:40:37 +0000 (22:40 -0700)
If this multiplication overflows:

usecs = (t * inv_cycles_per_usec) / 16777216UL;

then usecs is 2^64/2^24, which is 1099511627776. Divide that by
10^6 to get seconds, and that is 1099511. Since we cached the
old value previously, we'd get stuck with this amount of seconds.

To avoid turning this into an expensive division, have a check
and only divide if we have to. This avoids the overflow.

Signed-off-by: Jens Axboe <axboe@fb.com>
gettime.c

index e274671..30513e0 100644 (file)
--- a/gettime.c
+++ b/gettime.c
@@ -16,6 +16,7 @@
 #if defined(ARCH_HAVE_CPU_CLOCK) && !defined(ARCH_CPU_CLOCK_CYCLES_PER_USEC)
 static unsigned long cycles_per_usec;
 static unsigned long inv_cycles_per_usec;
+static uint64_t max_cycles_for_mult;
 #endif
 int tsc_reliable = 0;
 
@@ -182,7 +183,10 @@ static void __fio_gettime(struct timeval *tp)
 #ifdef ARCH_CPU_CLOCK_CYCLES_PER_USEC
                usecs = t / ARCH_CPU_CLOCK_CYCLES_PER_USEC;
 #else
-               usecs = (t * inv_cycles_per_usec) / 16777216UL;
+               if (t < max_cycles_for_mult)
+                       usecs = (t * inv_cycles_per_usec) / 16777216UL;
+               else
+                       usecs = t / cycles_per_usec;
 #endif
                tp->tv_sec = usecs / 1000000;
                tp->tv_usec = usecs % 1000000;
@@ -296,6 +300,7 @@ static int calibrate_cpu_clock(void)
 
        cycles_per_usec = avg;
        inv_cycles_per_usec = 16777216UL / cycles_per_usec;
+       max_cycles_for_mult = ~0ULL / inv_cycles_per_usec;
        dprint(FD_TIME, "inv_cycles_per_usec=%lu\n", inv_cycles_per_usec);
        return 0;
 }