Well, there's always
$year=int($year);

[download]

But Perl is not a strongly-typed language. You can also use
$year+=0;

[download]

to ensure that $year is interpreted as a number in following statements (as long as the following statements don't do anything to the variable).

However, the code you show should make year a number in any case. Perhaps the code actually is
$year = '1992';

[download]

or is otherwise assigned a value interpreted as a string.

In reply to Re: Type casting by HyperZonk
in thread Type casting by Anonymous Monk

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