After being baffled for minutes on end about the seeming madness of ++$X + $X++ being 3, I divined the reason:
  1. $X is 0
  2. ++$X increments $X's value, and returns the variable $X (not its value)
  3. $X++ return's $X's value and then increments it, meaning $X is now 2
  4. 2 + 1 = 3
This can be verified by the Perl debugger: 
perl -de0
main::(-e:1):   0
  DB<1> x $N = 0; ++$N, $N++
0  2
1  1

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_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker, who'd like a (from-home) job
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

In reply to Re: $i=$i++ by japhy
in thread $i=$i++ by Anonymous Monk

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