For the people interested in the solution: here is what I learned.
I was used to the compiler of  VW and the one of squeak is a shock:
menu built in parser, dependency from the parser to the UI, missing 
exception,
strange logic: a beast :)

So for my problem I added an OutOfScopeNotification signal but this is 
bad since this signal
will only be invoked in none interactive mode.

So I think that getting the new compiler in is really important.


Encoder>>undeclared: name

	| sym |
	OutOfScopeNotification signal.
	requestor interactive ifTrue: [
		requestor requestor == #error: ifTrue: [requestor error: 
'Undeclared'].
		^ self notify: 'Undeclared'].
	Transcript show: ' (' , name , ' is Undeclared) '.
	sym := name asSymbol.
	Undeclared at: sym put: nil.
	^self global: (Undeclared associationAt: sym) name: sym


On 17 févr. 05, at 15:02, stéphane ducasse wrote:

> Hi
>
> I would like to check whether certain string are regular Smalltalk 
> expressions (without evaluating them)
> So I played with the compiler and its lack of consistent exceptions. 
> Anyway.
>
> Compiler new compile: 'zorkgloubibolga123456789 ', 'go' in: Bot 
> notifying: nil ifFail: [nil]
> returns
>  zorkgloubibolga123456789
> 	
> 	go
> while I would like to be able to get an syntaxErrorNotification or 
> OutOfScopeNotification since go is not an instance variable of class 
> Bot.
>
>
> I tried
>
> [Compiler new compile: 'zorkgloubibolga123456789 ', 'go: 120' in: Bot 
> notifying: nil ifFail: [nil]]
> 		on: SyntaxErrorNotification
> 		do: [ Beeper beep]
> and there I get an exception which is good for me.
>
>
>
> Does anybody has an idea how I can get an exception on the following 
> case.
>
> [Compiler new compile: 'zorkgloubibolga123456789 ', 'go' in: Bot 
> notifying: nil ifFail: [nil]]
> 		on: OutOfScopeNotification
> 		do: [ Beeper beep]
>
>