Il 19/07/2013 08:48, Peter Lieven ha scritto:
>> @@ -3040,7 +3040,10 @@ int64_t bdrv_get_block_status(BlockDriverState
>> *bs, int64_t sector_num,
>> int coroutine_fn bdrv_is_allocated(BlockDriverState *bs, int64_t
>> sector_num,
>> int nb_sectors, int *pnum)
>> {
>> - return bdrv_get_block_status(bs, sector_num, nb_sectors, pnum);
>> + int64_t ret = bdrv_get_block_status(bs, sector_num, nb_sectors,
>> pnum);
>> + return
>> + (ret & BDRV_BLOCK_DATA) ||
>> + ((ret & BDRV_BLOCK_ZERO) && !bdrv_has_zero_init(bs));
>
> i do also not understand the "((ret & BDRV_BLOCK_ZERO) &&
> !bdrv_has_zero_init(bs))";
> if a block is unallocated and reads as zero, but the device lacks zero
> init, it is declared as allocated with this, isn't it?
Perhaps bdrv_has_zero_init(bs) could be replaced by bs->backing_hd. I'd
have to look at the code more closely.
But I suggest that you try using get_block_status (the function in
qemu-img.c) instead of bdrv_is_allocated in qemu-img.c. It will
probably simplify your code _and_ make it more efficient.
> for iscsi and host_device with lbprz==1 or discardzeroes respectively all
> blocks would return as allocated. is this wanted?
Paolo
