On Sat, Mar 8, 2014 at 1:44 PM, Scott dunning <swdunn...@cox.net> wrote: >> if 1 > guess > 100: >> > OH! I see what you're saying, ignore my last post. Yes that looks > cleaner.
Please read section 6.9 of the language reference, which defines Python comparison expressions. http://docs.python.org/3/reference/expressions#not-in Here's the description of chained comparisons: Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false). Formally, if a, b, c, ..., y, z are expressions and op1, op2, ..., opN are comparison operators, then a op1 b op2 c ... y opN z is equivalent to a op1 b and b op2 c and ... y opN z, except that each expression is evaluated at most once. Note that a op1 b op2 c doesn’t imply any kind of comparison between a and c, so that, e.g., x < y > z is perfectly legal (though perhaps not pretty). Thus `1 > guess > 100` is equivalent to `(guess < 1) and (guess > 100)`, which is always false. The correct chained comparison is `not (1 <= guess <= 100)`. Chaining is generally simpler, since all expressions are only evaluated once. In this particular case, with a local variable compared to constants, the chained form is slightly less efficient in CPython. Though if it "looks cleaner" to you, certainly use it. Readability takes precedence. _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
