[myhdl-list] another intbv question

[Neal B. <ndb...@gm...>]

I noticed that

x = intbv (5, -16, 15)

print x._min
> -16

print (x >> 1)._min
None

This is surprising.  I would have expected that arithmetic operations on 
intbv would result in an intbv with the same #bits.

I have implemented a fixed pt arithmetic class in c++ (based on 
boost::constrained_value).  What I did there is all operations result in the 
same # bits.  If you want something different, first convert operands to the 
desired result size.  For example, multiplying 2 8-bit values will give an 
8-bit result.  If you wanted a 16-bit result, first convert the inputs to the 
wider field.

Others have argued that these conversions should be automatic (e.g., 8 x 8 -
> 16), but I prefer explicit.

In any case, I think the intbv behavior is surprising.  Is this really 
desirable?