Hi folks,
I was working Exercise 6 in Chapter 5 of CTM, and I came across a couple of
issues.
First of all, I think the last *else* line in Figure 5.22 (p. 393) needs an
*end* after it.
Secondly, contrary to Exercise 6a, I do not think the second form reduces to
the third form when the time-out delay is zero. Here's some sample Erlang:
receive
{rectangle, [X,Y]} -> X*Y;
{circle, [R]} -> 3.14159*R*R;
after [ 0 -> 0 ];
end
Here's the Oz translation, using the second form (with some boilerplate
added):
declare Sin Result Sout
Sin=[foo rectangle([2 3])]
Result = local
Cancel={Alarm 0}
fun{Loop S T#E Sout}
if {WaitTwo S Cancel}==1 then
case S of M|S1 then
case M
of rectangle([X Y]) then E=S1 T=Sout X*Y
[] circle([R]) then E=S1 T=Sout 3.14159*R*R
else E1 in E=M|E1 {Loop S1 T#E1 Sout} end
end
else E=S T=Sout 0 end
end T
in
{Loop Sin T#T Sout}
end
{Browse Sin#Result#Sout}
Now here's the translation of the third form (with the same boilerplate):
declare Sin Result Sout
Sin=[foo rectangle([2 3])]
Result = if {IsDet Sin} then
case Sin of M|S1 then
case M
of rectangle([X Y]) then S1=Sout X*Y
[] circle([R]) then S1=Sout 3.14159*R*R
else Sin=Sout 0 end
end
else Sout=Sin end
{Browse Sin#Result#Sout}
The second form gives the result [foo rectangle([2 3])]#6#[foo], but the
third form gives the result [foo rectangle([2 3])#0#[foo rectangle([2 3])].
The problem is that the third form eliminates the loop that was present in
the other two forms. I'm not sure why that facet was eliminated.
Am I missing something?
Thanks,
Lyle
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