Re: $i=$i++

You have to understand how expressions evaluate. The pre- and post-increment (and decrement) operators change the variables value, but the variable plus the operator constitute and expression who's value varies depending upon the placement of the ++. Essentially, the value of the expression $i++ is original value of $i and $i is incremented by 1. The value of the expression ++$i is new value of $i after $i is incremented by 1.

$i = ++$i + $i++;
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In the above example, we evaluate the expression right to left. The expression, $i++, evaluates as zero, even though $i is now 1. The second expression, ++$i, evaluates as 2 (since $i is now 1) and 2+1 equals 3.

$i = $i++ + ++$i;
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Pretty straightforward now. The rightmost ++$i evaluates as 1 and $i is set to one. The $i++ also evaluates to 1 (remember, with a post-increment operator doesn't change the return value of the expression, just the variable -- confused yet? :). So, 1 + 1 equals two.

Cheers,
Ovid

Update: Got the concept right, but I evaluated in reverse order. Whoops! Juerd is right.

Update2: Okay, I'm smoking crack. If things are evaluated left to right, then the following is true:

$ perl -e 'print 1-3+2'
0
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Well, it turns out that this is true. If it were evaluated right to left, then the answer would negative four. What I think is going on (though I can't prove it offhand) is that built-in unary operators are evaluated right to left, and then the rest of the expression is evaluated based upon precedence rules, if any. Anyone want to shed light on this?

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Re: Re: $i=$i++ by Juerd (Abbot) on Apr 29, 2002 at 15:50 UTC
In the above example, we evaluate the expression right to left. Is that always, or just in this case? When are expressions evaluated left-to-right, and when right-to-left? And why is that so? I would have thought that `do { $i = 0; ++$i + $i++; }` would evaluate to 2 because (0 + 1) + 1 == 2, but apparently, I'm wrong. Update: Got the concept right, but I evaluated in reverse order. Whoops! Juerd is right. Answer to Ovid's update: No, I'm not, as do { $i = 0; ++$i + $i++ } does NOT equal 2 -- the right-to-left apparently is correct, I'm just wondering why it is that way. - Yes, I reinvent wheels. - Spam: Visit eurotraQ.	[reply] [d/l]
Re: Re: $i=$i++ by Juerd (Abbot) on Apr 29, 2002 at 17:04 UTC
If it were evaluated right to left, then the answer would negative four No, it's a per-operator thing, as + and - are different operators, and their precedence defines they should be evaluated left-to-right. This is done by implicit parentheses: `2;0 juerd@ouranos:~$ perl -MO=Deparse,-p -e'sub one { 1 } sub two { 2 +} sub three { 3 } print one() - three() + two()' sub one { 1; } sub two { 2; } sub three { 3; } print(((one() - three()) + two()));` [download] (I had to use these subs, because Perl optimizes 1-2+3 to a literal 0 :) So we have these calculation: 1 - 3 = -2 answer ( = -2 ) + 2 = 0 But which are evaluated first? the 1 or the 3? The -2 or the 2? It becomes clear when you test the operators one by one (or so I thought): `2;0 juerd@ouranos:~$ perl -le'sub foo { print "foo"; return 1 } sub ba +r { print "bar"; return 2 } print foo + bar' bar foo 1 2;0 juerd@ouranos:~$ perl -le'sub foo { print "foo"; return 1 } sub ba +r { print "bar"; return 2 } print foo && bar' foo bar 2` [download] So I decided to test a lot of operator this way, but found out that everything was evaluated left-to-right, but not if I left out the parens. Why did my first test print bar before foo then? Because I didn't print `foo() + bar()` but did print `foo( +bar() )`. So that still doesn't solve our mistery... - Yes, I reinvent wheels. - Spam: Visit eurotraQ.	[reply] [d/l] [select]