in reply to Re^3: return +0
in thread return +0

Here's a nice simple example of when you might need to use it: 

$ perl -e 'print (2+3)/5'
5

[download]

How on earth does that print 5? Because it evaluates as "print the result of 2 + 3, then divide print's return value (which is 1 if print worked) by 5, and then do nothing with the result". This is obvious if you stick another print in there: 

$ perl -e 'print print (2+3)/5'
50.2

[download]

which is parsed as:

print((print(2+3))/5)

Replies are listed 'Best First'.
Re^5: return +0
by goibhniu (Hermit) on Nov 27, 2007 at 18:31 UTC

    And just to close the logic loop (++DrHyde and lidden, btw), this leading plus helps the parser evaluate all those numbers into a single value before printing. 

    $ perl -e 'print +(2+3)/5'
1

    [download]

    So I assume the author of the code the OP was reading thought there might be some ambiguity and wanted to help the parse out with that return statement.

    I humbly seek wisdom.
[reply]
[d/l]
[select]

In Section Seekers of Perl Wisdom