12.12.2017 19:32, Eric Blake wrote:
On 12/11/2017 09:24 AM, Vladimir Sementsov-Ogievskiy wrote:
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;
Not the same. 'block_size' in your code is 'uint32_t', so your
multiplication is still 32-bit if you don't cast blocks; while qemu has::
include/block/block.h:#define BDRV_SECTOR_BITS 9
include/block/block.h:#define BDRV_SECTOR_SIZE (1ULL << BDRV_SECTOR_BITS)
and the 1ULL in the qemu definition forces 'unsigned long long' results
whether or not you cast.
Ah, cool, missed this. And we can't do such thing for BDRV_SECTOR_BITS
because it will be unspecified behavior.
Thank you for explanation.
--
Best regards,
Vladimir
