Re: [myhdl-list] intbv.signed()

[Jan D. <ja...@ja...>]

Günter Dannoritzer wrote:
> Jan Decaluwe wrote:
> ...
>> I would go for an integer to start with. That is what MyHDL does now for
>> many operators. It is more efficient than constructing intbv's (although
>> I dont' know how much more) and seems sufficient in many cases.
> 
> I was thinking about the processing of the signed() function and
> basically need to do a categorization when the value of an intbv
> instance is considered unsigned. In that case the the signed() function 
> will change the value based on the msb bit, which is considered the sign 
> bit. In the other case the value is considered signed and the function 
> will return it just as is.
> 
> When considering intbv instances there are several combinations of
> min/max values possible. The following diagram shows them:
> 
> 
>      ----+----+----+----+----+----+----+----
>         -3   -2   -1    0    1    2    3
> 
> 1                     min  max
> 2                          min  max
> 3                min       max
> 4                min            max
> 5           min            max
> 6           min       max
> 7           min  max
> 8   neither min nor max is set
> 9   only max is set
> 10  only min is set
> 
> 
> The horizontal line shows a number range from -3 to 3. Then each line 
> shows different cases of min/max values of an intbv instance.
> 
> As an example, in the first case min=0 and max > 0. The second case min 
>  > 0 and max > min, etc.
> 
>  From all those cases I think only two cases, namely 1 and 2, are where
> the value is considered unsigned and the signed() function will return a
> modified value.
> 
> In all other cases either there is no range specified or the min value
> is < 0, which makes that intbv instance already considered signed and
> the signed() function will just return the value as is.
> 
> So the test will be for min >= 0 and _nrbits > 0. That case will 
> classify the intbv instance as unsigned and the signed() function will 
> return the value as signed, with the sign being determined based on the 
> msb. The msb is bit # _nrbits-1.
> 
> Note that case #10 (only min is set) could fulfill the requirement min 
>  >=0, but as no max is set, _nrbits is 0 and hence the second test will 
> not be True.

Sounds all very logical to me.

Jan


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