I think the reason it works this way is simpler than you make out. The important thing is the implementation of the preincrement operator: do { $m = 20; ++$m } does not return the value 20, rather it returns the variable $m.

Therefore the sequence of operations reduces to:

  $m = 20;              # $m == 20
  ++$m;                 # $m == 21
  $temp1 = $m++         # $m == 22, $temp1 == 21
  $temp2 = $m + $temp1  # 22 + 21 = 43
  print $temp2
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Hope this helps,

Hugo