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Subject: Re: [PHP-DEV] [RFC] Disallow multiple default blocks in a single
 switch statement
From: rowan.collins@gmail.com (Rowan Collins)

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Ferenc Kovacs wrote (on 13/08/2014):
> I think you misunderstood that part, I was thinking about a state 
> machine like this:
>
>  1. function doStuff(){
>  2. switch($state){
>  3. case OPENDOOR:
>  4. if(!opendoor()){
>  5. break;
>  6.       }
>  7.     $state = SITDOWN;
>  8. case SITDOWN:
>  9. if(!sitdown()){
> 10. break;
> 11.       }
> 12.     $state = SIPWHISKEY;
> 13. case SIPWHISKEY:
> 14.       sipwhiskey();
> 15.   }
> 16. }
>
>
> where you modify the switch variable in one case and fall through into 
> another.
> as I mentioned I don't have a reasonable use-case for multiple 
> defaults, but I can see some for multiple case labels in general, and 
> I don't think that it is better to have different behavior for case 
> and default in this regard.


The assignments to $state in that code do not make any difference to 
execution, control simply flows forwards whenever there is no "break".

$state is only tested once, to select the initial label to jump to, so 
no amount of reassigning it, or adding duplicate labels, can ever cause 
a second jump.

I'm tempted to write a PHP script that emulates the switch execution by 
building a set of if and goto statements, which might make some of this 
behaviour clearer (not just for this discussion, but for other confusion 
I've seen elsewhere), but I don't have time right now.

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