This is odd.

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

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Re: This is odd. by almut (Canon) on May 22, 2008 at 23:47 UTC
This is a precedence problem. You need to put parentheses around the `($n1 + $n2)`. Otherwise, Perl adds the result of concatenating the string `"$n1 plus $n2 is "` with `$n1` — i.e.the string `"4 plus 2 is 4"` — with `$n2`. This string numerically evaluates to the initial 4, so you do get the result 6.	[reply] [d/l] [select]
Re: This is odd. by dragonchild (Archbishop) on May 23, 2008 at 00:12 UTC
+, -, and . are operators with the same precedence and they have left-associativity. Thus, your line is parsed: `print( ((("$n1 plus $n2 is " . $n1) + $n2) . "\n") );` [download] Read perlop for more info. My criteria for good software: Does it work? Can someone else come in, make a change, and be reasonably certain no bugs were introduced?	[reply] [d/l]
Re: This is odd. by hesco (Deacon) on May 23, 2008 at 05:10 UTC
And put one more way, if I may . . . The dot operator is used to concatenate string data into a single scalar. Once you've done that, you're trying to use a (+) operator to do arithmetic between a string scalar and an integer, which is not defined. You confuse the poor compiler this way. -- Hugh if( $lal && $lol ) { $life++; }	[reply]
Re^2: This is odd. by JDelmoso (Novice) on May 23, 2008 at 05:22 UTC
Oh, my. Poor compiler. What do the commas do, then? I don't know why I used them, probably picked them up at some point dabbling around in Perl earlier.	[reply]
Re^3: This is odd. by mr_mischief (Monsignor) on May 23, 2008 at 19:21 UTC
They make print aware that it's dealing with a list of different items, due to list operator precedence rules. According to perlop: In the absence of parentheses, the precedence of list operators such as print, sort, or chmod is either very high or very low depending on whether you are looking at the left side or the right side of the operator. For example, in `@ary = (1, 3, sort 4, 2); print @ary; # prints 1324` [download] the commas on the right of the sort are evaluated before the sort, but the commas on the left are evaluated after. In other words, list operators tend to gobble up all the arguments that follow them, and then act like a simple TERM with regard to the preceding expression. Note that you have to be careful with parentheses: `# These evaluate exit before doing the print: print($foo, exit); # Obviously not what you want. print $foo, exit; # Nor is this. # These do the print before evaluating exit: (print $foo), exit; # This is what you want. print($foo), exit; # Or this. print ($foo), exit; # Or even this.` [download]	[reply] [d/l] [select]
Re: This is odd. by casiano (Pilgrim) on May 23, 2008 at 06:51 UTC
The behavior is the correct one if you supress the warnings flag: `p2@nereida:~/src/perl/testing$ cat precedence1.pl my $n1 = 2; my $n2 = 4; print "$n1 plus $n2 is " . $n1 + $n2 . "\n";` [download] When you run this program there are no warnings: `pp2@nereida:~/src/perl/testing$ perl precedence1.pl 6` [download] However: `pp2@nereida:~/src/perl/testing$ cat precedence2.pl use warnings; my $n1 =2; my $n2 = 4; print "$n1 plus $n2 is " . $n1 + $n2 . "\n";` [download] Issues the warning: `pp2@nereida:~/src/perl/testing$ perl precedence2.pl Argument "2 plus 4 is 2" isn't numeric in addition (+) at precedence2. +pl line 6. 6` [download] The warning flag and the `Warnings` module are full of heuristics that helps you most of the time. Whenever you have problems with operator priorities use the `p` option of `Deparse` like this: `pp2@nereida:~/src/perl/testing$ perl -MO=Deparse,-p precedence2.pl use warnings; use strict 'refs'; (my $n1 = 2); (my $n2 = 4); print(((("$n1 plus $n2 is " . $n1) + $n2) . "\n")); precedence2.pl syntax OK` [download] It will tell you the execution order of the involved operations. Observe that the extraordinary thing is that the operation `(((("$n1 plus $n2 is " . $n1) + $n2) . "\n"))` [download] still gives 6!. This is because the phrase starts with `4 plus ...`. There is some irony here. The first time I stomped in this sort of "Perl priority problems" I was puzzled. But if you live with it time enough you start to appreciate the fun of it Hope it helps Casiano	[reply] [d/l] [select]
Re: This is odd. by hesco (Deacon) on May 23, 2008 at 06:05 UTC
Your commas delimit the elements of a list, which are printed, one after another. In your use of the commas, the precendence is respected in a way that it does what you mean here. everything between the commas is a single element of the list or array, and must be evaluated itself, before the list is put together. That is different that the dot concatenation string operator. -- Hugh if( $lal && $lol ) { $life++; }	[reply]
Re^2: This is odd. by JDelmoso (Novice) on May 23, 2008 at 06:18 UTC
Ooh, excellent. Thanks so much.	[reply]
Re: This is odd. by roboticus (Chancellor) on May 23, 2008 at 11:26 UTC
JDelmoso: Not quite ... an odd number is one having a nonzero remainder after dividing by two. Two, four and six are all even! </tongue_in_cheek> ...roboticus	[reply]
Re: This is odd. by JDelmoso (Novice) on May 23, 2008 at 00:19 UTC
Oh, I see. Thanks.	[reply]