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Re: [Qemu-block] [PATCH v6 09/20] parallels: Switch to .bdrv_co_block_st

From: Vladimir Sementsov-Ogievskiy
Subject: Re: [Qemu-block] [PATCH v6 09/20] parallels: Switch to .bdrv_co_block_status()
Date: Mon, 11 Dec 2017 18:24:48 +0300
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:52.0) Gecko/20100101 Thunderbird/52.5.0

09.12.2017 19:39, Eric Blake wrote:
On 12/09/2017 06:31 AM, Vladimir Sementsov-Ogievskiy wrote:
07.12.2017 23:30, Eric Blake wrote:
We are gradually moving away from sector-based interfaces, towards
byte-based.  Update the parallels driver accordingly.  Note that
the internal function block_status() is still sector-based, because
it is still in use by other sector-based functions; but that's okay
because request_alignment is 512 as a result of those functions.
For now, no optimizations are added based on the mapping hint.

Signed-off-by: Eric Blake <address@hidden>

       BDRVParallelsState *s = bs->opaque;
-    int64_t offset;
+    int count;

+    assert(QEMU_IS_ALIGNED(offset | bytes, BDRV_SECTOR_SIZE));
-    offset = block_status(s, sector_num, nb_sectors, pnum);
+    offset = block_status(s, offset >> BDRV_SECTOR_BITS,
+                          bytes >> BDRV_SECTOR_BITS, &count);

+    *pnum = count * BDRV_SECTOR_SIZE;
       if (offset < 0) {
           return 0;

+    *map = offset;
oh, didn't notice previous time :(, should be

*map = offset * BDRV_SECTOR_SIZE

Oh, right.

(also, if you already use ">> BDRV_SECTOR_BITS" in this function,
would not it be more consistent to use "<< BDRV_SECTOR_BITS"
for sector-to-byte conversion?)

with that fixed (with "<<" or "*", up to you),
'>> BDRV_SECTOR_BITS' always works.  You could also write '/
BDRV_SECTOR_SIZE' (the compiler should figure out that you are dividing
by a power of 2, and use the shift instruction under the hood), but I
find that a bit harder to reason about.

On the other hand, '<< BDRV_SECTOR_BITS' only produces the same size
output as the input; if the left side is 32 bits, it risks overflowing.
But '* BDRV_SECTOR_SIZE' always produces a 64-bit value.  So I've
learned (from past mistakes in other byte-conversion series) that the
multiply form is less likely to introduce unintended truncation bugs.

hm, now I doubt. I've wrote tiny program:

#include <stdint.h>
#include <stdio.h>

int main() {
    uint32_t blocks = 1 << 20;
    int block_bits = 15;
    uint32_t block_size = 1 << block_bits;
    uint64_t a = blocks * block_size;
    uint64_t b = blocks << block_bits;
    uint64_t c = (uint64_t)blocks * block_size;
    uint64_t d = (uint64_t)blocks << block_bits;
    printf("%lx %lx %lx %lx\n", a, b, c, d);
    return 0;

and it prints
0 0 800000000 800000000
for me. So, no difference between * and <<...

Reviewed-by: Vladimir Sementsov-Ogievskiy <address@hidden>

Best regards,

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