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

[download]

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

[download]

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

— Ken

In reply to Re: Multiple numeric not or compare in if statement by kcott
in thread Multiple numeric not or compare in if statement by RedJeep

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