Actually, I get "FirstSecond2311".

The extra 1s are the return values from the functions first and second. As they don't have explicit return statements, they both return the result of the last expression evaluated. In both cases, that's a call to print. And print returns 1 (i.e. true) if it prints successfully.

For extra credit, explain why it doesn't print "23First1Second1" - which would seem to be the most obvious result.

Actually, I get "FirstSecond2311".

$ perl -le 'print 0'
0
$ perl -le 'print print 0'
0
1
$ perl -le 'print print print 0'
0
1
1
$ perl -le 'print print print print 0'
0
1
1
1
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Hint: print does have a return value!

For extra credit, explain why it doesn't print "23First1Second1" - which would seem to be the most obvious result.

Because the arguments to print are all evaluated first, then printed. ie, it's effectively:

@args = (2,3,first,second);
print @args;
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So 'first' and 'second' are run (generating output), before the print command w/ the list is performed.

Similarly, foreach (&func1, &func2) { ... } will run both 'func1' and 'func2' before starting the loop:

Read more... (494 Bytes)

Yes, FirstSecond2311. Me, too.
So THAT is where the extra 1s come from!

To rephrase what blazar, jhourcle and friedo (and the book) pointed out,
the terms in the main print statements have a higher precedence then
the plain numbers, and hence get evaluated first. The
said "terms" are subroutine calls that themselves do
a print, and that is why we see First and then Second being printed.
Thanks for the extra credit question, too, as well as
the answer.

To comment on adrianh suggesting other books instead,
the fact is I've read a ton of perl books. The
debugging book just happens to be one I haven't
read. In spite of all the Perl books I've read, the
I could not seem to figure out the answer.
Thanks again

FirstSecond2311
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That's because the function print returns 1 on success, just like many other Perl functions. Lacking an explicit return, statement, the first and second subs return the result of their last statement, which is 1 in both cases.

I am reading DeBUGGING PERL by Martin Brown

Personally I'd recommend putting it in the bin and going and buying a good book on Perl instead :-) It's one I would never recommend. Try the Llama book, or Peter Scott's Perl Debugged.

So where does that last digit '1' come from? Why isn't it just

I think you'll find that there are two "1"s at the end. They are the result of calling the first() and second() subroutines. If you don't have a explicit return the last value returned in the subroutine is returned. The print in first() and second() both return "1" after a successful print. So:

becomes