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Re: Multiple numeric not or compare in if statement

by kcott (Bishop)
on Jul 02, 2020 at 05:42 UTC ( #11118792=note: print w/replies, xml ) Need Help??


in reply to Multiple numeric not or compare in if statement

G'day RedJeep,

There's short-circuiting occurring which I think may be throwing you. (77 is not equal to 3) is TRUE; so, when $a is 77, ($a != 3) is TRUE. Having found a TRUE condition, there's no need to evaluate the other ORed conditions (so Perl doesn't).

Changing your Option Not Equal condition to completely negate the Option Equal condition will do what I think you want:

$ perl -E '$a = 77; if (($a == 3) || ($a == 77) || ($a == 8)) { say "i +s" } else { say "not" }' is $ perl -E '$a = 77; if (!(($a == 3) || ($a == 77) || ($a == 8))) { say + "not" } else { say "is" }' is

You can run that rather complicated condition through B::Deparse to find a simpler way of expressing it.

$ perl -MO=Deparse -e '(!(($a != 3) || ($a != 77) || ($a != 8)))' not $a != 3 || $a != 77 || $a != 8; -e syntax OK $ perl -E '$a = 77; if (not $a != 3 || $a != 77 || $a != 8) { say "not +" } else { say "is" }' is

If it helps you, perlop has an "Operator Precedence and Associativity" section.

— Ken

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