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Andrew Welch wrote:
Thanks for the correction. But now we have duplicated logic (the cast + ends-with), which some consider bad programming practice. Here's an update that does not duplicate the logic, removes the nested for (better: it shows a way of shortcutting nested loops with the comma operator), corrects my error and is much less readable (which was not on purpose, btw). for $i in reverse(1 to 99), $j in (1 to 3) return concat ($i - xs:integer($j mod 3 = 0), ' bottle', ('s')[not($i * 3 - $j = (1 to 3))], ' of beer', (' on the wall.', '. Take one down, pass it around', ' on the wall. ' )[$j] (I like the idea of using the predicate instead of if...then...else) As a tutorial it suits well for explaining why: ('s')[$i * 3 - $j != (1 to 3)] yields exactly the same result as: ('s')[$i * 3 - $j = (1 to 3)] which is quite counter-intuitive (IIRC, for a '!=' to return false, none from the left must be the same as any from the right; to be true, only one item needs to be unequal. This is not backed up by the example above, but I am sure I am overlooking something. Moreover, I found that (1 to 2) != 10 returns false, and (1,2) != 10 returns true.... I am really missing something here, this must be a faq somewhere :S ).
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