|We don't bite newbies here... much|
Untillian Headache or about the semantic of untilby Discipulus (Monsignor)
|on Feb 11, 2013 at 13:50 UTC||Need Help??|
I want share an headache with you due to the Perl's loop control until structure.
While explaining iterators apllied to permutations the author, with the usual kindness, come back a little to present a simpler problem: an odometer.
Realized that the odometer is that set of little wheels showing how many kilometers my kawasaki GPZ 750 divoured in his history, i looked at the subroutine proposed with a "ah ok... is very simple. i can get it.." approach.
This line, clear in the result wanted, mazed and puzzled me:
This is the original code of the sub with a little prepended code just to call it, putted by me.
The semantic translatedNow come in count another think, i'm not eng-native (as my name suggest i speak latin..).
Every word in english have an attached logical translation in my mind and, while i read some perl code, this is used as logical unit to build up the scenario: i cannot think in english..
I know that until simply invert the condition of the same while loop.
The while in my native language is mentre and the reverse until is 'finché'.
But while the first transaltion fit well in any circumstance, the second does not apart from trivial case of the condition.
I took the english-italian dictionary and i found that until is translated as finché (non)'. This beacuse in english until want a positive sentence but in italian 'finché bring the possibility to use a negative form of the sentence after it, and in this way is commonly used: i wait until you arrive can be translated as aspetto finché arrivi but also sounds good as 'aspetto finché NON arrivi'.
In italian the double negation is admitted and correct: Non c'è nessuno is There is nobody but have a double negation ('non' and 'nessuno') but sound as 'there is not nobody'.
Inverting the conditionNaturally the original code is correct (this was never in doubt) but now i want a valid logical representation of it in my mind so i played with the line with the until condition:
It does not appear the first attempt using an LABEL if redo solution because i think i'm too young to use LABELs.
But really the negation of ($odometer[$wheel] < 9 || $wheel < 0) is ($odometer[$wheel] == 9 && $wheel >= 0) ?
UPDATE 10 Nov 2017 found a similar article at blogs.perl.org and left here my comments and a link to this post.
there are no rules, there are no thumbs..
Reinvent the wheel. Then learn The Wheel and use it!