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Re^2: Identical Arrays

by Anonymous Monk
on Aug 26, 2012 at 05:07 UTC ( #989786=note: print w/ replies, xml ) Need Help??


in reply to Re: Identical Arrays
in thread Identical Arrays

# the sort order (numeric or alpha) doesn't matter # as long as it is consistent.
my @x = ('1ringy-dingy2ringy-dingy', 2, 3); my @y = (2, 3, 1);


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Re^3: Identical Arrays
by Marshall (Prior) on Aug 26, 2012 at 05:30 UTC
    Well, indeed!

    I've never used smart match in production code because as this example confirms, it is not as "smart" as one might think! I stand corrected about this. I guess you have to very smart to use the "smart match". Obviously, I am not that smart.

      It's a can of worms. :)

      Here's what's happening:

      First, when you perform @arrayA ~~ @arrayB the action you get is actually similar to ($arrayA[0] ~~ $arrayB[0]) && ($arrayA[1] ~~ $arrayB[1]) && .... So you're really getting a chain of scalar to scalar smart-matches.

      Next we have to scan down the list to find something that fits the description of 'scalar ~~ scalar', while keeping in mind that the left-hand side looks like a string, and the right hand side looks like a number. Here's the obscure rule:

      Any ~~ Num: numeric equality like: Any == Num

      (~~ added for clarity)

      So you're comparing something like this: '1ringy-dingy2ringy-dingy' ~~ 1. The right-hand side is a number. So a numeric comparison will be performed. And what happens when you use "1ringy-dingy2ringy-dingy" as a number? Well, if a string starts with a number, Perl will use that number and drop the rest when performing numeric operations. So '1ringy-dingy2ringy-dingy' ~~ 1 is the same as '1ringy-dingy2ringy-dingy' == 1, which is the same as 1 == 1.


      Dave

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