Re: [PATCH v2 for-4.7 07/14] libxc: fix uninitialized variable

[Wei Liu <wei.liu2@citrix.com>]

On Wed, Apr 27, 2016 at 10:57:17AM +0200, Roger Pau Monne wrote:
> On Tue, Apr 26, 2016 at 04:16:51PM +0100, Wei Liu wrote:
> > On Tue, Apr 26, 2016 at 04:52:16PM +0200, Roger Pau Monne wrote:
> > > Signed-off-by: Roger Pau Monné <roger.pau@citrix.com>
> > > ---
> > > Cc: Ian Jackson <ian.jackson@eu.citrix.com>
> > > Cc: Wei Liu <wei.liu2@citrix.com>
> > > ---
> > >  tools/libxc/xc_dom_bzimageloader.c | 2 +-
> > >  1 file changed, 1 insertion(+), 1 deletion(-)
> > > 
> > > diff --git a/tools/libxc/xc_dom_bzimageloader.c b/tools/libxc/xc_dom_bzimageloader.c
> > > index 7fde42a..0a4041c 100644
> > > --- a/tools/libxc/xc_dom_bzimageloader.c
> > > +++ b/tools/libxc/xc_dom_bzimageloader.c
> > > @@ -482,7 +482,7 @@ static int xc_try_lzo1x_decode(
> > >          if ( !dst_len )
> > >          {
> > >              msg = "Error registering stream output";
> > > -            if ( xc_dom_register_external(dom, out_buf, out_len) )
> > > +            if ( xc_dom_register_external(dom, out_buf, 0) )
> > 
> > I fail to figure out why this is correct.
> > 
> > I think the out_len should be moved out of the loop and initialise as 0.
> > We still need to use out_len here.
> 
> I'm not following here. AFAICT out_len is always uninitialized at this 
> point (from loop 0 to N), so I assume 0 is what was intended to be passed 
> here. Or it this supposed to be passing in the last out_len value from the 
> previous iteration of the loop?
> 

Note that out_buf is defined out of the scope of this loop, which points
to the buffer returns to caller when decompression completes (note the
return 0 in that block).  The second argument to
xc_dom_register_external should reflect the true size of the buffer.

I got confused by the name of out_len because it is actually used for
something else. So please ignore what I said in my earlier email about
using out_len. From reading the code I think we need to use *size or
something. Does this make sense?

Wei.

> Roger.

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