in reply to precedence question

try: 
#!/usr/bin/perl
$a = 10;
$a+=5;$a-=10;
print "\n$a";

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or 
#!/usr/bin/perl
$a = 10;
$a = $a + 5 - 10;
print "\n$a";

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The += and -= are shortcuts that set the lefthand.. the $a = is not needed.

-Waswas
Edited because I was speaking jibberish at the bottom (heh, talking on the phone while typing =)

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Re: Re: precedence question
by Anonymous Monk on Aug 09, 2002 at 20:45 UTC
    First of all, sorry about the additional post; refreshing the page on my browser had that unfortunate side effect. However, disregarding where I put the result, suppose I changed it to : 
    #!/usr/bin/perl
$a = 10;
$result = ($a+=5) + ($a-=10);
print "$result";

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    My question is, since the assignment is made to the scalar variable $a using += and -= at supposedly different times, which parentheses is evaluated first? Or are they evaluated at the same time? I know there are different ways to do this; this is a kind of "What is the philosophy of Perl on this situation?" type of question, because I was curious. The possible answers I can see are:
    1: parenthesis right to left: (0+5)+(10-10) = 5.
    2: parenthesis left to right: (10+5)+(15-10) = 20.
    3: parenthesis simultaneously: (10+5)+(10-10) = 15.
    However, when this is run, with the modifications above, I still get 10 as an answer.
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[d/l]
      Take this example: 
      $a = 30;
$one = ($a += 5) + ($a -= 10);
print "a=$a one= $one \n";

      [download]
      Output: a=25 one=50 it looks the the += is being updated after the calculation: so (30) + (30 - 10) = 50 then after the line passes you see $a is 25. Not sure if this is how it should work or not seems kinda fishy that the += is not proccessed the same way -= is. but i still don't understand what you are trying to add up here? $result = ($a+=5) + ($a-=10); is not very intuitive. Do you expect $result to be the result status of the += or -=? I am not sure what I would even expect $result to hold += and -= are not really normal arithmatic operators...

      -Waswas
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[d/l]
[select]
        The way it was explained to me, the value of $a is modified with all the parenthetical calls, and then the result is added. So the actual calculation would end up being:
        $a = 30
        then $a = 35 (after the +=5)
        then $a = 25 (after the -=10)
        then $one = $a + $a, or (25)+(25) = 50, our answer.
        I knew the code wasn't intuitive; it was presented as a specifically nonintuitive snippet in order to show the problems inherent in unclear code, in the example I was looking at. I got curious as to the result, and fretted over it until Zaxo laid it out for me.
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