Don't think so. My interpretation of his code and his question is that he wants to use symbolic references: See "I have several arrays in my script" and "with $ARGV[1] as the array *name*". So in $arr1 is the name of the real array and he wants to access that.
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You're quite likely correct; it's just that from the code fragment given by the OP, I couldn't find where his array was being populated. I also tend to think that an event like $#arr1 giving an incorrect value is unlikely; I would expect that $#arr1 is among the Perl features least likely to have any bugginess. My thinking (I'm easily confused by symbolic references, so I'm likely wrong) is that $#arr1 isn't quite the same as scalar(@$arr1)-1.
Information about American English usage here and here. Floating point issues? Please read this before posting. — emc
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Exactly, if there is a @arr1 and you access it without any symbolic reference, then $#arr1 is scalar(@arr1)-1. Only @arr1 is accessed here, the value of $arr1 is irrelevant and isn't used at all
If we use $arr1 (totally different variable from @arr1) as a symbolic reference, then
@bingo=(1,2,3);
$arr1='bingo';
print $#{$arr1}, scalar(@$arr1)-1, scalar(@bingo)-1;
would print 222, as all three print expressions would be equivalent. In this case @arr1 is irrelevant and not used at all, only $arr1 as the symbolic reference and @bingo as the array
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