On Mon, Feb 08, 2010 at 12:47:50AM +0100, Ronny Klier wrote:
> I think there is a bug in label encoding in Format6Encoder. For some
> string length the last encoded byte is not stored.
>
> E.g. having a string "10007" the encoded byte buffer looks like this
>
> [0] [0x86]
> [1] [0x8]
> [2] [0x20]
> [3] [0x9f]
> [4] [0xf0]
>
> The number of stored bytes is 4. So the 0xf0 will not show up in the
> final image file.
> Index: Format6Encoder.java
> ===================================================================
> --- Format6Encoder.java (Revision 1541)
> +++ Format6Encoder.java (Arbeitskopie)
> @@ -86,7 +86,7 @@
>
> buf = put6(buf, off++, 0xff);
>
> - int len = ((off - 1) * 6) / 8 + 1;
> + int len = (int)Math.ceil((off * 6) / 8.0);
You can do this with integer math, truncating division.
Your example was off=6 (5 chars and the end-of-string code),
and I suppose we would get len=4 instead of 5:
(6-1)*6 / 8 + 1 = 30/8 + 1 = 3.75 + 1 = 4
If you want to round up to full blocks, the normal trick is to
add divisor-1 before dividing, like this:
int len = ((off - 1) * 6 + 7) / 8 + 1 = 4.625 + 1 = 5
I don't know if the off-1 and the +1 are correct. An integer version
of your formula would also work in this case:
int len = (off * 6 + 7) / 8 = 43/8 = 5.375 = 5
This formula is clear to me: it will clearly convert the "off" 6-byte chars
(including the end-of-string code) to the number of required 8-bit octets.
Best regards,
Marko
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