Re: [myhdl-list] Slicing an unsigned intbv to a signed one

[Christopher L. F. <cf...@uc...>]

Yes, that is a much better solution.  Implementing something similar  
appears to achieve the goals.  As before I did not implement anything  
for the slicing to test.  The slicing approach would be similar (and  
testing probably pretty weak, need to test other bit widths, unsigned,  
etc.).

------ Example Output ------
 >>> x = intbv(7, min=-8, max=8)
 >>> x
intbv(7)
 >>> str(x)
'0111'
 >>> x[3] = 1
 >>> x
intbv(-1L)
 >>> str(x)
'1111'
 >>> x[0] = 0
 >>> x
intbv(-2L)
 >>> str(x)
'1110'
 >>> x[1] = 0
 >>> x
intbv(-4L)
 >>> str(x)
'1100'
 >>> x[2] = 0
 >>> x
intbv(-8L)
 >>> str(x)
'1000'
 >>> x[3] = 0
 >>> x
intbv(0L)
 >>> str(x)
'0000'
 >>> x[3] = 1
 >>> x
intbv(-8L)
 >>> str(x)
'1000'
 >>> x[:] = 3
 >>> x
intbv(3L)
 >>> str(x)
'0011'
 >>> x[3] = 1
 >>> x
intbv(-5L)
 >>> str(x)
'1011'
 >>>

----- Diff ------

154,163d153
<
<              
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<             # CFELTON EDIT
<             # Create a sign bit mask
<             if self._min < 0 and key == self._nrbits-1:
<                 smask = (-1 << self._nrbits-1)
<             else:
<                 smask = 0
<              
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
<
165c155
<                 self._val |= ((1L << i) | smask)
---
 >                 self._val |= (1L << i)
167,169c157
<                 self._val &= (~(1L << i) & ~smask)
<
<
---
 >                 self._val &= ~(1L << i)
196d183
<
449,455c436
<         s = ''
<         for i in range(self._nrbits):
<             if self._val & (1 << i):
<                 s = '1' + s
<             else:
<                 s = '0' + s
<         return s #str(self._val)
---
 >         return str(self._val)


On Jun 13, 2008, at 12:20 PM, Günter Dannoritzer wrote:

> Christopher L. Felton wrote:
>> Ok, below was my quick test to double check if it could be done.  A
>> real implementation probably need more thought (if feasible at all).
> ...
>
> I think what you need to do is not just multiply by -1, but create a
> mask that is ORed with the actual number. If the sign bit gets set,  
> the
> mask will be all 1 upwards from the sign bit and all 0 down to bit 0.
> Then OR that with the current intbv value.
>
> So instead of this:
>
>   self._val = -1 * self._val
>
> do this:
>
>   self._val = (-1 << self._nrbits-2) | self._val
>
> This is basically a merging between the existing bits below the sign  
> bit
> and the sign change to negative.
>
> For clearing the sign bit the simplest way would be to slice the bits
> below the sign bit out and return them as a new intbv value.
>
> Guenter
>
>
>
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