>
> R. Joseph Newton wrote:
> > Mark VanMiddlesworth wrote:
> >> What's wrong with the syntax, especially line 12?
>
> >> if ($arg = "-n") {$e = 0} # Line 12
Sorry if I'm misunderstanding what your're tryin gto do but to me it seems :
$arg = $ARGV[0]; # or whatever/however you assign the value of '-n' to $arg
if($arg eq "-n") { ...
Or
if($arg =~ m/^\-n$/) { ...
If you are simplt checking to see if $arg has a value of '-n' then the = is not right,
just on equal sign is assigning a value and then you're getting different results than
if you were doing a comparison operator like 'eq', 'ne' or '==', '!='.
Etc.
So if you're wanting to check $arg's value then you need to use a comparison operator
not an assignment one.
> >
> > You are asking, in the line above, whether your assignment
> of "-n" to
> > the scalar $arg succeeded. Answer: true, always.
>
> Not quite. Mark is asking if the value contained in $arg is
> 'true' after assigning "-n" to it. As everyone knows, N is
> always true, while Y is often false. This is preferable:
>
> if ($arg = -x) {$f = 0} # Line 13
>
> as it is only true in the rarest of cases, and won't
> spuriously set your $f variable to zero without a
> second thought.
>
> ;-)
>
> Rob
>
>
>
>
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