It's because unlike += the ++ perl operator (prefix or postfix) returns an rvalue, not an lvalue. You can't do ++$a = 8 or $a++ = 8 either.

Update: swapped rvalue and lvalue, as noted by calin.

Perl follows the C convention for the increment/decrement operators, or nearly so. ++$foo++ is undefined because only one *-ment op is allowed per variable in a statement. Would you like that to do (++$foo)++, or ++($foo++) ? It would make a difference, but there is nothing in the grammar to distinguish them.

(++$foo)++ and ++($foo++) don't work either, for reasons mentioned by ambrus.

I was thinking about that this week, because I wanted (for no good reason), things like:

$n+++;        # increment, then do that again (add 2)
$n++++;       # increment, then again, and once more (add 3)
$n++++++++++; # increment N-1 times, just because I can

--$n++; # use the value that's one less, then restore it
++$n--; # same thing, other way

$n++-;   # increment, but not really cause this is a JAPH
$n++-+;  # I changed my mind, really do it this time
$n++-+-; # boss says change it back 

$n+-     # can I decide later, or, pick one at random

$n**     # no exponent, so square $n
$n***    # cube it, and so on

$n***/   # square it, but cube it first and go back one
$n//     # 1
$n///    # 1/$n
$n///**  # $n
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and that would be horribly annoying. Instead of  $n += 5; people would write $n++++++... does that seem pretty? On the other hand, it wouldn't be all that hard to write a source filter to do this...

If you want +++ operator, rather use them for addition with lower precedence, for example this

$a + $b ** $c + $d ++ $e + $f ** $g + $h *** 
        $i + $j ** $k + $l ++ $m + $n ** $o + $p
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would become

(($a + $b) * ($c + $d) + ($e + $f) * ($g + $h)) * 
        (($i + $j) * ($k + $l) + ($m + $n) * ($o + $p))
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    ($a+=1)++
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