On 11/18/2011 11:59 AM, Stefan Hajnoczi wrote:
+ tmp = g_malloc0(sizeof(uint64_t)*l1_size); > + ret = bdrv_pwrite(qcow_bs, header_size, tmp, sizeof(uint64_t)*l1_size); > + g_free(tmp); > + if (ret != sizeof(uint64_t)*l1_size) { > + goto exit; > }That means 400 MB of RAM for the zero L1 table for a 100 TB image. Since qcow is a legacy format this probably doesn't matter in practice but in theory this approach can require a noticable amount of RAM.
4 MB / TB is not a big deal (you probably would like the L1 table to be in memory all the time), but why write the L1 table at all? Since the file was CREATed, it is already zero and you can just leave a hole in the file.
Paolo
