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

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On Wed, 2009-10-07 at 08:54 -0400, tedd wrote:

> At 6:15 PM -0700 10/6/09, Daevid Vincent wrote:
> >Except that:
> >
> >$a = 123;
> >$b = $a++;
> >echo $b;  //gives 123, not 124
> >
> >as you logically expect it to and common sense would dictate, regardless of
> >what K&R or anyone else says.
> 
> That's not the way I look at it.
> 
>     $b = $a++;
> 
> means to me "take the value of $a and assign to $b and then increment $a."
> 
> Whereas:
> 
>     $b = ++$a;
> 
> means to me "increment $a and take the value of $a and assign to $b."
> 
> Cheers,
> 
> tedd
> 
> -- 
> -----
> http://sperling.com  http://ancientstones.com  http://earthstones.com
> 


Which is exactly the reason for the two operators in C.

Thanks,
Ash
http://www.ashleysheridan.co.uk



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