Jeff Shannon wrote:
if myrange in range(10,90): # "in" is the key word here return True else return False
This is, however, the correct solution. :)
Or I *should* say, rather, that this is *a* correct solution, in that it will yield the expected answer. Kent Johnson's '10 < x < 90' is a better solution, however -- 'if x in range(...)' creates a list of numbers, and then steps through that list comparing x to each one in turn, while Kent's version makes only two comparisons and no object creations. While one should never prematurely optimize, it's also good to be aware of how much work is done by different options, and in this case 'if x in range()' isn't any clearer than the direct comparisons. I'd think nothing of the extra cost of using range() if it *did* make the code easier to read, but there's no sense in going to extra work for no benefit. :)
Jeff Shannon Technician/Programmer Credit International
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