Re: Whacky increment/assignment logic with $foo++ vs ++$foo

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At 10:48 AM -0300 10/6/09, Martin Scotta wrote:
No matter how silly it can looks like (a = a++) it is still completely valid
code and it SHOULD run without problems.

Yeah, it's a valid as:

   $a = $a;

and does the same thing, which is nothing.

If you want a statement that does something, then use:

  $a = ++$a;

or simply:

  $a++;

or

  ++$a;

Any of those will increment $a, whereas ($a = $a++;) does nothing.

Cheers,

tedd

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