On Tue, 12/02 12:10, Don Koch wrote:
> On Tue, 2 Dec 2014 15:39:14 +0800
> Fam Zheng <f...@redhat.com> wrote:
>
> > Zeroing a buffer that will be filled right after is not necessary, and
> > allocating a power of two + 1 is naughty.
> >
> > Suggested-by: Markus Armbruster <arm...@redhat.com>
> > Signed-off-by: Fam Zheng <f...@redhat.com>
> > ---
> > block/vmdk.c | 5 +++--
> > 1 file changed, 3 insertions(+), 2 deletions(-)
> >
> > diff --git a/block/vmdk.c b/block/vmdk.c
> > index 28d22db..0c5769c 100644
> > --- a/block/vmdk.c
> > +++ b/block/vmdk.c
> > @@ -558,14 +558,15 @@ static char *vmdk_read_desc(BlockDriverState *file,
> > uint64_t desc_offset,
> > }
> >
> > size = MIN(size, 1 << 20); /* avoid unbounded allocation */
> > - buf = g_malloc0(size + 1);
> > + buf = g_malloc(size);
>
> For a file that is less than 1<<20 bytes, won't this throw away the last byte?
> Maybe better is:
> size = MIN(size, (1 << 20) - 1);
> buf = g_malloc(size + 1);
> Leave the bdrv_pread as it was and...
>
> >
> > - ret = bdrv_pread(file, desc_offset, buf, size);
> > + ret = bdrv_pread(file, desc_offset, buf, size - 1);
> > if (ret < 0) {
> > error_setg_errno(errp, -ret, "Could not read from file");
> > g_free(buf);
> > return NULL;
> > }
> > + buf[ret - 1] = 0;
>
> ...zero the last byte changes to:
> buf[ret] = 0;
Yes. I'll respin this one. Thanks.
Fam
>
> >
> > return buf;
> > }
> > --
> > 1.9.3
>
> -d
