note 93847 deleted from language.operators.increment by danbrown

Note Submitter: daevid at daevid dot com ---- <?php $num = 123; $num = $num++; print $num; //this prints 123 and not 124 ?!! $num = 123; $num = ++$num; print $num; //this prints 124 as expected $num = 123; $num++; print $num; //this prints 124 as expected $num = 123; print $num++; //this prints 123 and not 124 ?!! print $num++; //this NOW prints 124 ?> So then I read the manual because I think I'm loosing my mind and perhaps it's backwards day and nobody told me. I'm baffled as to the reasoning behind: "$a++ :: Post-increment :: Returns $a, then increments $a by one." Why would you EVER want $num = $num++; to give you back the value you already had? Even if we did $foo = $bar++; I would still logically (and common sensely) expect $foo to be the increment of $bar! It also seems counter-intuitive, as I was always lead to believe everything is processed on the right of an equals sign and then assigned back to the left side of the equals sign. In this case, it does a mixture of both.