ping: fix standard deviation for small, consistent ping times

Suppose we do 10 pings, and 5 take 10us, 5 take 11us.  tsum2 is 1105, tsum
is 105.  We divide them, getting 110 and 10 respectively.  tmdev is then
llsqrt(110 - 10 * 10) = llsqrt(10) = 3us.  But max-min is only 1us!

The exact solution is sqrt(110.5 - 110.25) which is sqrt(0.25) = 0.5.  The
new code will (more) correctly compute llsqrt(0) = 0 as the answer.  It
takes care not to do any horrendous integer overflows if there's a lot of
long pings (a few thousand >1s pings would overflow without this).

Addresses: https://bugs.debian.org/907327
Signed-off-by: Sami Kerola <kerolasa@iki.fi>

ping_common.c

@@ -956,14 +956,22 @@ void finish(void)
 
 	if (nreceived && timing) {
 		double tmdev;
+		long total = nreceived + nrepeats;
+		long tmavg = tsum / total;
+		long long tmvar;
+
+		if (tsum < INT_MAX)
+			/* This slightly clumsy computation order is important to avoid
+			 * integer rounding errors for small ping times. */
+			tmvar = (tsum2 - ((tsum * tsum) / total)) / total;
+		else
+			tmvar = (tsum2 / total) - (tmavg * tmavg);
 
-		tsum /= nreceived + nrepeats;
-		tsum2 /= nreceived + nrepeats;
-		tmdev = llsqrt(tsum2 - tsum * tsum);
+		tmdev = llsqrt(tmvar);
 
 		printf(_("rtt min/avg/max/mdev = %ld.%03ld/%lu.%03ld/%ld.%03ld/%ld.%03ld ms"),
 		       (long)tmin/1000, (long)tmin%1000,
-		       (unsigned long)(tsum/1000), (long)tsum%1000,
+		       (unsigned long)(tmavg/1000), (long)(tmavg%1000),
 		       (long)tmax/1000, (long)tmax%1000,
 		       (long)tmdev/1000, (long)tmdev%1000
 		       );