Re^4: Pre vs Post Incrementing variables

[repellent]

$i = 0; ++$i = 'fred';;
Can't modify preincrement (++) in scalar assignment


So, not an lvalue.

Hmm. I was looking at this prior to my previous post. Notice that:

do { my $i = 16; ${ \(++$i) } = 'fred'; $i };  # returns 'fred'

do { my $i = 16; ${ \($i++) } = 'fred'; $i };  # returns 17

Maybe the warning you got is due to how Perl parses the preincrement assignment expression? I don't know.

[BrowserUk]

Waddayaknow! And I've even found a use for it:

print map ${ \( ++$i ) } %= 10, 1..30;;
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0

Now all we've gotta do is document it, and we've a valid reason for it working that way: measuring golf games. Perfect :)

[BrowserUk]

Mind you, it might make explaining this one a tad awkward :)

$i = 0; print ++${\++$i}, 0+ ++$i, 0+ ++$i, ++${\++$i};;
6 3 4 6

[repellent]

The argument list to print is evaluated from left-to-right. ++${\++$i} is merely incrementing $i twice and returning an alias to $i.


$i = 0; print ++${\++$i}, 0+ ++$i, 0+ ++$i, ++${\++$i};;
              ^              ^        ^     ^
             (1)            (2)      (3)   (4)


(1) & (4) return aliases to $i. Since $i has been incremented six times, we get 6 for both of them.

(2) & (3) return copies to $i because of the zero additions 0+. We get 3 for (2) after three increments, and 4 for (3) with one more increment.

[BrowserUk]

You appear to be saying that the ability to describe what the implementation does, justifies why it does it that way.

The fact that this:

$i=0; print $n = ++$i, $m = ++$i;;
1 2

produces a different result to this:

$i=0; print ++$i, ++$i;;
2 2

Is just plain weird. Some, including me, would say 'broken'.

Justifying the weirdness by saying "that's what it does", kinda reminds of Apple's "You're holding it wrong." explanation of Antennagate.