On Fri, Dec 30, 2016, at 09:47, Steve D'Aprano wrote:
> Again, assume both operands are in range for an N-bit signed integer.
> What's
> a good way to efficiently, or at least not too inefficiently, do the
> calculations in Python?
I'd do something like:
bit_mask = (1 << bits) - 1 # 0xFFFF
sign_bit = 1 << (bits - 1) # 0x8000
sign_ext = ~bit_mask # ...FFFFF0000
def signed(value):
if value & sign_bit:
return value | sign_ext
else:
return value & bit_mask
def unsigned(value):
return value & bit_mask
And also avoid doing it on intermediate steps where it can be shown to
not affect the result or allow the value to grow too large.
--
https://mail.python.org/mailman/listinfo/python-list
