21 May
2018
21 May
'18
1:17 p.m.
On Fri, May 18, 2018 at 09:47:53AM +0800, Marek Lindner wrote:
@@ -251,6 +253,21 @@ static void
batadv_tp_caller_notify(struct batadv_priv *bat_priv,
break;
case BATADV_TP_ELP:
+ if (reason_is_error) {
+ batadv_v_elp_tp_fail(tp_vars->hardif_neigh);
+ return;
+ }
+
+ test_time = jiffies_to_msecs(jiffies - tp_vars->start_time);
+ total_bytes = atomic64_read(&tp_vars->tot_sent);
+
+ /* The following calculation includes these steps:
+ * - convert bytes to bits
+ * - divide bits by the test length (msecs)
+ * - convert result from bits/ms to 0.1Mb/s (* 1024 * 10 / 1000)
+ */
+ throughput = total_bytes * 8 >> ilog2(test_time) / 10;
+ batadv_v_elp_tp_finish(tp_vars->hardif_neigh, throughput);
I find the throughput calculation quite hard to read here, would
it be possible to put this into an extra (inline?) function?
Also the comment for the "convert result..." seems wrong,
"[bits/ms]*1024*10/1000" would be 0.01Mb/s, not 0.1Mb/s?
What is the advantage of using the ilog2 and shift operator here
compared to plain multiplications and divisions?
Also, when trying this in a small C program I get weird results:
-----
#include <stdio.h>
#include <math.h>
int main()
{
unsigned long test_time = 10000; // 10s
unsigned long total_bytes = 20000000; // 16MBit/s
unsigned long throughput, throughput2;
unsigned long log_test_time = log(test_time) / log(2);
throughput = total_bytes * 8 >> log_test_time / 10;
// Straightforward approach?
throughput2 = total_bytes * 8 / test_time * 1000 / 1024 / 100;
printf("Result: %lu (log_test_time: %lu)\n", throughput,
log_test_time);
printf("Result2: %lu\n", throughput2);
return 0;
}
-----
$ ./test
Result: 80000000 (log_test_time: 13)
Result2: 156
$ file ./test
./test: ELF 32-bit LSB pie executable ARM, EABI5 version 1 (SYSV), dynamically linked,
interpreter /lib/ld-linux-armhf.so.3, for GNU/Linux 3.2.0,
BuildID[sha1]=d18f32829cdd2bc42cf744cdcafde7cdbd315cb0, not stripped
-----
Regards, Linus