Does Python need a '>>>' operator?

[David Eppstein]

In article <a9dbc9$bge$0 at 216.39.172.122>, bokr at oz.net (Bengt Richter) 
wrote:

> All 32-bit ints are represented in hex without a sign, so

This signed-unsigned confusion bit me when I needed a routine to count 
the nonzero bits of some input.  Normally one would write

    def popCount(x):
        count = 0
        while x:
            count += 1
            x &= x - 1

but of course this gets into an infinite loop when the high bit is set.  
Arguably, this is the correct behavior: a negative number in 2's 
complement should be thought of as having an infinite sequence of 1-bits.

But in my case, my input didn't ever have the high bit set, and a good 
thing too, or I would have gotten ValueError when I tried 
int(hexstring,16).
The reason I was getting into trouble anyway was because of the way 
right-shifting can change sign:

>>> 0x40000000<<1
-2147483648

This is inconsistent with int-long unification.  In fact, I think it is 
a bug.  The answer should, of course, be 2147483648L, just as you would 
get by multiplying by 2 instead of shifting.

-- 
David Eppstein       UC Irvine Dept. of Information & Computer Science
eppstein at ics.uci.edu http://www.ics.uci.edu/~eppstein/