SF bug 594996: OverflowError in random.randrange

Loosened the acceptable 'start' and 'stop' arguments so that any
Python (bounded) ints can be used.  So, e.g., randrange(-sys.maxint-1,
sys.maxint) no longer blows up.
This commit is contained in:
Tim Peters 2002-08-16 03:41:39 +00:00
parent 012c0a393a
commit 9146f27b77
1 changed files with 17 additions and 4 deletions

View File

@ -75,6 +75,7 @@ used to "move backward in time":
from math import log as _log, exp as _exp, pi as _pi, e as _e
from math import sqrt as _sqrt, acos as _acos, cos as _cos, sin as _sin
from math import floor as _floor
__all__ = ["Random","seed","random","uniform","randint","choice",
"randrange","shuffle","normalvariate","lognormvariate",
@ -299,7 +300,7 @@ class Random:
"""
# This code is a bit messy to make it fast for the
# common case while still doing adequate error checking
# common case while still doing adequate error checking.
istart = int(start)
if istart != start:
raise ValueError, "non-integer arg 1 for randrange()"
@ -307,14 +308,26 @@ class Random:
if istart > 0:
return int(self.random() * istart)
raise ValueError, "empty range for randrange()"
# stop argument supplied.
istop = int(stop)
if istop != stop:
raise ValueError, "non-integer stop for randrange()"
if step == 1 and istart < istop:
try:
return istart + int(self.random()*(istop - istart))
except OverflowError:
# This can happen if istop-istart > sys.maxint + 1, and
# multiplying by random() doesn't reduce it to something
# <= sys.maxint. We know that the overall result fits
# in an int, and can still do it correctly via math.floor().
# But that adds another function call, so for speed we
# avoided that whenever possible.
return int(istart + _floor(self.random()*(istop - istart)))
if step == 1:
if istart < istop:
return istart + int(self.random() *
(istop - istart))
raise ValueError, "empty range for randrange()"
# Non-unit step argument supplied.
istep = int(step)
if istep != step:
raise ValueError, "non-integer step for randrange()"