Case semantic

[Enrique Iurleo]

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.

[Sébastien Doeraene]

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

Post by Enrique Iurleo
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
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.

[Enrique Iurleo]

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?

Post by Sébastien Doeraene
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

Post by Enrique Iurleo
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
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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http://www.mozart-oz.org/mailman/listinfo/mozart-users

[Sébastien Doeraene]

I'm not sure I understand what you mean by "ASA". But if I do, then yes,
you're right.
Sébastien

Post by Enrique Iurleo
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?

[Enrique Iurleo]

Sorry haha.. i mean the Store (the SAS)

Post by Sébastien Doeraene
I'm not sure I understand what you mean by "ASA". But if I do, then yes,
you're right.
Sébastien

Post by Enrique Iurleo
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?

_________________________________________________________________________________
mozart-users mailing list
http://www.mozart-oz.org/mailman/listinfo/mozart-users

[Sébastien Doeraene]

OK, then you were right, indeed. The store is unchanged.
Sébastien

Post by Enrique Iurleo
Sorry haha.. i mean the Store (the SAS)