[Libre-soc-bugs] [Bug 784] Implement cl* instructions for carry-less operations

[bugzilla-daemon at libre-soc.org]

https://bugs.libre-soc.org/show_bug.cgi?id=784

--- Comment #18 from Luke Kenneth Casson Leighton <lkcl at lkcl.net> ---
(In reply to Jacob Lifshay from comment #13)
> (In reply to Luke Kenneth Casson Leighton from comment #5)
> >    comb += addend1.eq(a ^ b)
> >    comb += addend2.eq(a | b)
> 
> I explicitly have it more complex so I can explain all the cases and to
> match 1:1 with the reference algorithm.

that could be achieved with a comment (not too big! the code's really
elegant and small without the case statement.  something like:

"""
for i in range(self.width):
   for each bit a[i] b[i] set addend1[i], addend2[i]
      case 1,1     both have no leading zeros so far, so set carry 1,1
      case 0/1 1/0  different number of leading zeros, so clear carry 0,0
      case 0,0      propagate results from lower bits, so set 1,0

this is basically equivalent to: addend1 = a^b and addend2 = a|b
"""

does that look reasonably obvious?

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