Martin Brown book page 33

mnooning has asked for the wisdom of the Perl Monks concerning the following question:

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Re: Martin Brown book page 33 by davorg (Chancellor) on May 16, 2006 at 10:05 UTC
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. -- <http://dave.org.uk> "The first rule of Perl club is you do not talk about Perl club." -- Chip Salzenberg	[reply]
Re^2: Martin Brown book page 33 by blazar (Canon) on May 16, 2006 at 11:19 UTC
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` [download] Hint: print does have a return value!	[reply] [d/l]
Re^2: Martin Brown book page 33 by jhourcle (Prior) on May 16, 2006 at 15:29 UTC
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;` [download] 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... (448 Bytes)	[reply] [d/l] [select]
Re^2: Martin Brown book page 33 by mnooning (Beadle) on May 17, 2006 at 02:37 UTC
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	[reply]
Re: Martin Brown book page 33 by adrianh (Chancellor) on May 16, 2006 at 10:07 UTC
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: `print(2,3,first,second);` becomes `print(2,3,1,1);`	[reply] [d/l] [select]
Re: Martin Brown book page 33 by friedo (Prior) on May 16, 2006 at 10:06 UTC
There's actually two extra 1's on there, the result is `FirstSecond2311` [download] 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.	[reply] [d/l] [select]