[myhdl-list] intbv with min -1, max 1 has 2 bits?

[Günter D. <dan...@we...>]

Hi,

I tried to create a signed intbv with one bit in the form:

 >>>a = intbv(-1,-1,1)
 >>>print "a: %d, min: %d, max: %d, _nrbits: %d"%(
           a, a.min, a.max, a._nrbits)

a: -1, min: -1, max: 1, _nrbits: 2

It turns out that intbv uses 2 bits for it. Looking in the 
intbv.__init__() function there is this code section starting at line 46:

            if max is not None and min is not None:
                 _nrbits = maxfunc(len(bin(max-1)), len(bin(min)))
                 if min >= 0:
                     _nrbits = len(bin(max-1))
                 elif max <= 0:
                     _nrbits = len(bin(min))
                 else:
                     # make sure there is a leading zero bit in positive 
numbers
                     _nrbits = maxfunc(len(bin(max-1))+1, len(bin(min)))


What is interesting about this is that the first assignment to _nrbits 
right after the if clause is actually overwritten by any of the 
following assignments.

So the reason for the -1,1 range to have 2 bits seems to lay in the 
comment about having a leading zero bit for positive numbers.

I might be oblivious and the reason is obvious, but I don't see it at 
the moment. Anyone can explain why I need a leading zero bit and cannot 
create a 1 bit intbv?

Cheers,

Guenter