Re: $ARGV dilemma

in reply to $ARGV dilemma

UPDATE: Err, I answered my own question right after submission; I apologize for asking such a silly question, and jumping to ask, before giving it any thought.
The solution is, of course, to just use unless $ARGV[2].

UPDATE 2: I guess I should have been more precise in both my question and update. Originally, the last argument was 0; but changing it to a string was the solution I came up with..

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Re: Re: $ARGV dilemma by dws (Chancellor) on May 27, 2001 at 07:42 UTC
`unless $ARGV[2]` works until someone passes a zero. You could amend it to `unless defined $ARGV[2]` though `if 3 > @ARGV` reads better, at least in my opinion.	[reply] [d/l] [select]
Re: Re: $ARGV dilemma by wog (Curate) on May 27, 2001 at 07:41 UTC
That won't always work. What if `@ARGV = ("0","0","0")`. Then `$ARGV[2]` is false and you get an error despite having three arguments. You probably like something like `unless @ARGV == 3`.	[reply] [d/l] [select]

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