qemu-devel
[Top][All Lists]
Advanced

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: [Qemu-devel] [PATCH v1 00/17] dataplane: optimization and multi virt


From: Ming Lei
Subject: Re: [Qemu-devel] [PATCH v1 00/17] dataplane: optimization and multi virtqueue support
Date: Sun, 10 Aug 2014 11:46:24 +0800

Hi Kevin, Paolo, Stefan and all,


On Wed, 6 Aug 2014 10:48:55 +0200
Kevin Wolf <address@hidden> wrote:

> Am 06.08.2014 um 07:33 hat Ming Lei geschrieben:

> 
> Anyhow, the coroutine version of your benchmark is buggy, it leaks all
> coroutines instead of exiting them, so it can't make any use of the
> coroutine pool. On my laptop, I get this (where fixed coroutine is a
> version that simply removes the yield at the end):
> 
>                 | bypass        | fixed coro    | buggy coro
> ----------------+---------------+---------------+--------------
> time            | 1.09s         | 1.10s         | 1.62s
> L1-dcache-loads | 921,836,360   | 932,781,747   | 1,298,067,438
> insns per cycle | 2.39          | 2.39          | 1.90
> 
> Begs the question whether you see a similar effect on a real qemu and
> the coroutine pool is still not big enough? With correct use of
> coroutines, the difference seems to be barely measurable even without
> any I/O involved.

Now I fixes the coroutine leak bug, and previous crypt bench is a bit high
loading, and cause operations per sec very low(~40K/sec), finally I write a new
and simple one which can generate hundreds of kilo operations per sec and
the number should match with some fast storage devices, and it does show there
is not small effect from coroutine.

Extremely if just getppid() syscall is run in each iteration, with using 
coroutine,
only 3M operations/sec can be got, and without using coroutine, the number can
reach 16M/sec, and there is more than 4 times difference!!!

From another file read bench which is the default one:

      just doing open(file), read(fd, buf in stack, 512), sum and close() in 
each iteration

without using coroutine, operations per second can increase ~20% compared
with using coroutine. If reading 1024 bytes each time, the number still can
increase ~10%. The operations per second level is between 200K~400K per
sec which should match the IOPS in dataplane test, and the tests are
done in my lenovo T410 notepad(CPU: 2.6GHz, dual core, four threads). 

When reading 8192 and more bytes each time, the difference between using
coroutine and not can't be observed obviously.

Surely, the test result should depend on how fast the machine is, but even
for fast machine, I guess the similar result still can be observed by
decreasing read bytes each time.


diff --git a/qemu-img-cmds.hx b/qemu-img-cmds.hx
index ae64b3d..78c3b60 100644
--- a/qemu-img-cmds.hx
+++ b/qemu-img-cmds.hx
@@ -15,6 +15,12 @@ STEXI
 @item bench [-q] [-f @var{fmt]} [-n] [-t @var{cache}] filename
 ETEXI
 
+DEF("co_bench", co_bench,
+    "co_bench -c count -f read_file_name -s read_size -q -b")
+STEXI
address@hidden co_bench [-c @var{count}] [-f @var{filename}] [-s 
@var{read_size}] [-b] [-q]
+ETEXI
+
 DEF("check", img_check,
     "check [-q] [-f fmt] [--output=ofmt]  [-r [leaks | all]] filename")
 STEXI
diff --git a/qemu-img.c b/qemu-img.c
index 3e1b7c4..c9c7ac3 100644
--- a/qemu-img.c
+++ b/qemu-img.c
@@ -366,6 +366,138 @@ static int add_old_style_options(const char *fmt, 
QemuOpts *opts,
     return 0;
 }
 
+struct co_data {
+    const char *file_name;
+    unsigned long sum;
+    int read_size;
+    bool bypass;
+};
+
+static unsigned long file_bench(struct co_data *co)
+{
+    const int size = co->read_size;
+    int fd = open(co->file_name, O_RDONLY);
+    char buf[size];
+    int len, i;
+    unsigned long sum = 0;
+
+    if (fd < 0) {
+        perror("open file failed\n");
+        exit(-1);
+    }
+
+    /* the 1st page should have been in page cache, needn't worry about block 
*/
+    len = read(fd, buf, size);
+    if (len != size) {
+        perror("open file failed\n");
+        exit(-1);
+    }
+    close(fd);
+
+    for (i = 0; i < len; i++) {
+        sum += buf[i];
+    }
+
+    return sum;
+}
+
+static void syscall_bench(void *opaque)
+{
+    struct co_data *data = opaque;
+
+#if 0
+    /*
+     * Doing getppid() only will show operations per sec may increase 5
+     * times in my T410 notepad via bypassing coroutine!!!
+     */
+    data->sum += getppid();
+#else
+    /*
+     * open, read 1024 bytes, and close will show ~10% increase in my
+     * T410 notepad via bypassing coroutine!!!
+     *
+     * open, read 512bytes, and close will show ~20% increase in my
+     * T410 notepad via bypassing coroutine!!!
+     *
+     * Below link provides 'perf stat' on several hw events:
+     *
+     *       http://pastebin.com/5s750m8C
+     *
+     * And with bypassing coroutine, dcache loads decreases, insns per
+     * cycle increased 0.7, branch-misses ratio decreases 0.4%, and
+     * dTLB-loads decreases too.
+     */
+    data->sum += file_bench(data);
+#endif
+
+    if (!data->bypass) {
+        qemu_coroutine_yield();
+    }
+}
+
+static int co_bench(int argc, char **argv)
+{
+    int c;
+    unsigned long cnt = 1;
+    int num = 1;
+    unsigned long i;
+    struct co_data data = {
+        .file_name = argv[-1],
+        .sum = 0,
+        .read_size = 1024,
+        .bypass = false,
+    };
+    Coroutine *co, *last_co = NULL;
+    struct timeval t1, t2;
+    unsigned long tv = 0;
+
+    for (;;) {
+        c = getopt(argc, argv, "bc:s:f:");
+        if (c == -1) {
+            break;
+        }
+        switch (c) {
+        case 'b':
+            data.bypass = true;
+            break;
+        case 'c':
+            num = atoi(optarg);
+            break;
+        case 's':
+            data.read_size = atoi(optarg);
+            break;
+        case 'f':
+            data.file_name = optarg;
+            break;
+        }
+    }
+
+    printf("%s: iterations %d, bypass: %s, file %s, read_size: %d\n",
+           __func__, num,
+           data.bypass ? "yes" : "no",
+           data.file_name, data.read_size);
+    gettimeofday(&t1, NULL);
+    for (i = 0; i < num * cnt; i++) {
+        if (!data.bypass) {
+            if (last_co) {
+                qemu_coroutine_enter(last_co, NULL);
+            }
+            co = qemu_coroutine_create(syscall_bench);
+            last_co = co;
+            qemu_coroutine_enter(co, &data);
+        } else {
+            syscall_bench(&data);
+        }
+    }
+    gettimeofday(&t2, NULL);
+    tv = (t2.tv_sec - t1.tv_sec) * 1000000 +
+        (t2.tv_usec - t1.tv_usec);
+    printf("\ttotal time: %lums, %5.0fK ops per sec\n", tv / 1000,
+           (double)((cnt * num * 1000) / tv));
+
+    return (int)data.sum;
+}
+
 static int img_create(int argc, char **argv)
 {
     int c;


Thanks,
-- 
Ming Lei



reply via email to

[Prev in Thread] Current Thread [Next in Thread]