Pre-increment *and* Post-increment

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

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Re: Pre-increment and Post-increment by ambrus (Abbot) on Feb 01, 2005 at 17:48 UTC
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.	[reply] [d/l] [select]
Re: Pre-increment and Post-increment by Zaxo (Archbishop) on Feb 01, 2005 at 17:50 UTC
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 `++($foo++)` ? It would make a difference, but there is nothing in the grammar to distinguish them. After Compline, Zaxo	[reply] [d/l] [select]
Re^2: Pre-increment and Post-increment by ikegami (Patriarch) on Feb 01, 2005 at 19:31 UTC
`(++$foo)++` and `++($foo++)` don't work either, for reasons mentioned by ambrus.	[reply] [d/l] [select]
Re: Pre-increment and Post-increment by brian_d_foy (Abbot) on Feb 01, 2005 at 20:20 UTC
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/// # $n [download] Surely Perl can be even more write-only if we just applied ourselves. -- brian d foy <bdfoy@cpan.org>	[reply] [d/l]
Re^2: Pre-increment and Post-increment by Anonymous Monk on Feb 01, 2005 at 23:20 UTC
and that would be horribly annoying. Instead of `$n += 5;` people would write `$n++++++...` does that seem pretty? On the other hand, it wouldn't be all that hard to write a source filter to do this...	[reply] [d/l] [select]
Re^2: Pre-increment and Post-increment by ambrus (Abbot) on Feb 03, 2005 at 20:57 UTC
If you want `+++` operator, rather use them for addition with lower precedence, for example this `$a + $b $c + $d ++ $e + $f $g + $h * $i + $j $k + $l ++ $m + $n ** $o + $p` [download] would become `(($a + $b) * ($c + $d) + ($e + $f) * ($g + $h)) * (($i + $j) * ($k + $l) + ($m + $n) * ($o + $p))` [download]	[reply] [d/l] [select]
Re: Pre-increment and Post-increment by holli (Abbot) on Feb 01, 2005 at 17:49 UTC
Why in the world do you want to do that? holli, regexed monk	[reply]
Re: Pre-increment and Post-increment by Anonymous Monk on Feb 02, 2005 at 09:18 UTC
You can't do `++$a++` because `++` returns a value, and not something you can modify. But you can write: `($a+=1)++` [download] which should have the same effect.	[reply] [d/l] [select]
Re: Pre-increment and Post-increment by tinita (Parson) on Feb 03, 2005 at 14:09 UTC
how about: `${ \++$a }++;` you can even do this as often as you want: `${ \++${ \++$a } }++;` =)	[reply] [d/l] [select]