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. -- Eric Blake, Principal Software Engineer Red Hat, Inc. +1-919-301-3266 Virtualization: qemu.org | libvirt.org
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