On 8/8/07, strk <[EMAIL PROTECTED]> wrote: > > On Thu, Aug 09, 2007 at 05:47:28AM +0200, strk wrote: > > On Wed, Aug 08, 2007 at 06:03:37PM -0700, David Rorex wrote: > > > On 8/8/07, Paul Smedley <[EMAIL PROTECTED]> wrote: > > > > > Why is it even calling ceil on an int in the first place? I think you > should > > > be able to simply do: (regardless of platform) > > > > > > in->ensureBytes( sample_count * ( 3 + n_bits/8 ); > > > > It's not calling ceil on an int, but on the result of a division. > > We want to get 1 if n_bits is 6 ... > > > > ceil(n_bits/8) > > I committed a patch fixing a bug in that function. > The new code still contains the ceil call, Paul, try adding a final .0 > to the constant in the division (4096.0 and 8.0) in case it fails the > way it is. > > --strk; >
Yeah, I was going to say that, if n_bits is an int, and 8 is an int, then the result will also be an int (as if floor had been called on it). adding the .0 will force floating point division.
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