On Mar 5, 6:34 pm, Gary Herron <gher...@islandtraining.com> wrote: > Mensanator wrote: > > On Mar 5, 3:42 pm, Gary Herron <gher...@islandtraining.com> wrote: > > >> Mensanator wrote: > > >>> The only way to get a 0 from a reverse range() is to have a bound of > >>> -1. > > >> Not quite. An empty second bound goes all the way to the zero index: > > > Not the same thing. You're using the bounds of the slice index. > > I was refering to the bounds of the range() function. > > >>>> for a in range(9,-9,-1):print(a,end=' ') > > > 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 > > > To get that to stop at 0, you use a -1 as the bounds: > > >>>> for a in range(9,-1,-1):print(a,end=' ') > > > 9 8 7 6 5 4 3 2 1 0 > > > Your slice notation only works if the last (first?) number in > > the range happens to be 0. What if the range bounds were variables? > > You may still want to force the range's last number to be 0 by > > using a constant like range(a,-1,-1) rather than just take > > the last number of range(a,b,-1) by using slice notation. > > All true and valid of course, but I was just contridicting the "the > ONLY way to get a 0" (emphasis mine) part of the statement.
Does it still contradict if you do not use the '::' as the OP requested? > > Gary Herron > > > > > > >> >>> range(9)[2::-1] > >> [2, 1, 0] > > >> Gary Herron- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - -- http://mail.python.org/mailman/listinfo/python-list
