Leo,
I've been away from email for the weekend, so sorry for the delay.
Leopold Toetsch wrote:
Well C and C<$I13> or some such is the same for the compiler. Both
need a Parrot register allocation. If they aren't reused after this
instruction, there isn't any problem to put these into the same
re
Nick Glencross <[EMAIL PROTECTED]> wrote:
> The problem is that $I13 has become *I16*, not (say) I17, so the line
> $I13 = word & 0xff00
> has actually had the side effect of changing 'word'.
Well C and C<$I13> or some such is the same for the compiler. Both
need a Parrot register allocatio
Leopold Toetsch wrote:
Well. That's it. Just use $I10 ... $I13 and they will stay what they
were. Put a print thereafter for debugging which uses these regs.
I should have emphasised the oddity a bit better.
'word' has mapped to I16 correctly, and I17 has been used for
$I10..$I12. That's fin
Nick Glencross <[EMAIL PROTECTED]> wrote:
> $I10 = word & 0x00ff
> $I11 = word & 0xff00
> $I12 = word & 0x00ff
> $I13 = word & 0xff00
> and the resulting code becomes
> band I17, I16, 0x00ff
> band I17, I16, 0xff00
> band
At 7:02 PM +0100 4/23/04, Nick Glencross wrote:
Ok, so doing a 'parrot -d 8' shows that word doesn't seem to
interfere with $I13 as far as it is concerned ...
Dumping the Interf. graph:
---
...
word -> $I10 $I11 $I12 (3)
...
$I10 -> word (1)
$I11 -> word (1
Ok, so doing a 'parrot -d 8' shows that word doesn't seem to interfere with
$I13 as far as it is concerned ...
Dumping the Interf. graph:
---
...
word -> $I10 $I11 $I12 (3)
...
$I10 -> word (1)
$I11 -> word (1)
$I12 -> word (1)
$I13 -> (0)
...
I won
I've now reduced the code down to this, a much simpler example [don't expect
it to run!]
Regards,
Nick
.sub _md5_create_buffer
.param string str
.local int word
$I0 = length str
$I1 = $I0 / 64
$I2 = $I0 % 64
$I3 = 64 * $I1
word = 0
md5_create_buffer_loop:
Guys,
I'll log a bug for this if need be, but I'd rather solve the problem myself
(with some help). I'm in the middle of writing a routine (MD5 hashing for
fun), and stumbled upon the following code generation problem in imcc.
In the attached code, I've got
$I10 = word & 0x00ff
$
