Yes i understand. So the ASA dont change? i mean only change the
environment right? so in my example Y bound to x.1 in the ASA.. am i right?
2012/4/11 Sébastien Doeraene <sjrdoera...@gmail.com>
> Hi,
>
> No, the case does not do a unification. Indeed, the argument X of the case
> is never modified/bound, even partially. However, non-escaped variable
> names in the *pattern* do indeed get bound to the corresponding parts in
> X (and these variables only belong to the environment in the "then" part of
> the given guard).
>
> Does that answer your question?
>
> Sébastien
>
>
> On Wed, Apr 11, 2012 at 00:58, Enrique Iurleo <quiquewol...@gmail.com>wrote:
>
>> Hi list, i have a doubt about the semantic of case. When i execute a case
>> the system does a unification of the variable and pattern? i mean.. for
>> example i have:
>>
>> local X in
>> X=[1 2 3]
>> case X of Y|Yr then {Browse Y} {Browse Yr} else skip end
>> end
>>
>> In this case when the "Browse" is executed browse "1" and "[2 3]".. so
>> the environment changes during the case? does the case a unification?
>>
>> Thanks i hope your answers.
>>
>
>
>
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