should this work? (it doesn't now)

|void main()
|{
|	foo: while(true)
|		break foo;   // works
|
|	switch(6)
|	{
|		case 5:
|			while(true)
|				break case 5;   // fails
|
|		case 6:
|			goto case 5;  // works
|
|	}
|}

"BCS" <ao@pathlink.com> wrote in message news:55391cb32812d8ca2d7783d0f566@news.digitalmars.com...
> should this work? (it doesn't now)
>
> |void main()
> |{
> | foo: while(true)
> | break foo;   // works
> |
> | switch(6)
> | {
> | case 5:
> | while(true)
> | break case 5;   // fails
> |
> | case 6:
> | goto case 5;  // works
> |
> | }
> |}

It's not in the grammar at all, so I'm not surprised that it doesn't.  It's a tiny feature though.

Reply to Jarrett,

> "BCS" <ao@pathlink.com> wrote in message
> news:55391cb32812d8ca2d7783d0f566@news.digitalmars.com...
> 
>> should this work? (it doesn't now)
>> 
>> |void main()
>> |{
>> | foo: while(true)
>> | break foo;   // works
>> |
>> | switch(6)
>> | {
>> | case 5:
>> | while(true)
>> | break case 5;   // fails
>> |
>> | case 6:
>> | goto case 5;  // works
>> |
>> | }
>> |}
> It's not in the grammar at all, so I'm not surprised that it doesn't.
> It's a tiny feature though.
> 

OK, I miss phrased the question: should this be in the grammer?

"BCS" <ao@pathlink.com> wrote in message news:55391cb3281338ca2d803bbfda0e@news.digitalmars.com...
>
> OK, I miss phrased the question: should this be in the grammer?

If you're proposing a feature, go ahead and propose it.  I mean, I can't see what I'd use this for, but..

Shouldn't the code be:

|void main()
|{
|foo: while(true)
|break foo;   // works
|
|bar: switch(6)
|{
|case 5:
|while(true)
|break bar;   // fails
|
|case 6:
|goto case 5;  // works
|
|}
|}

With embedded switches, "case 5" isn't always unique.  Also, the "break case 5" syntax almost implies to me that it should return to case 6.  I hope that's not what you wanted...

Reply to Jason,

> Shouldn't the code be:
> 
> |void main()
> |{
> |foo: while(true)
> |break foo;   // works
> |
> |bar: switch(6)
> |{
> |case 5:
> |while(true)
> |break bar;   // fails
> |
> |case 6:
> |goto case 5;  // works
> |
> |}
> |}
> With embedded switches, "case 5" isn't always unique.

Ok, but that same problem happens with goto case #;

based on that, this should be added:

outer: switch(n)
{
 case 1:
   switch(m)
   {
      case 2: goto outer case 2;
   }
 case 2:
}

>  Also, the
> "break case 5" syntax almost implies to me that it should return to
> case 6.  I hope that's not what you wanted...
> 

Ok so it loops forever, but I /was/ intending that the break exit the loop and continue down the case statment

This would be a more sane example. 

|void main()
|{
| foo: while(true)
|  break foo;   // works
|
| switch(6)
| {
|  case 5:
|   while(true)
|    break case 5;   // fails
|   somthing();
|   break;
|
|  case 6:
|   goto case 5;  // works
|
| }
|}

The point is that case ###: lables act like normal lables elseware, why not with break?