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<p>Hello,<br>
<br>
I've just started looking at XOTcl over the last couple days, and so far I like it a lot. Since I'm still learning, I assume I've overlooked something in the documentation, but I can't find how to remove a forwarded method from an object. It seems that all I can do is add a forwarded method, but once installed, I can never unforward a method. Is this true?<br>
<br>
Here's an (simplified) example of what I'm trying to do:<br>
<br>
Object a<br>
a proc foo {args} {<br>
puts "a: [self] foo"<br>
}<br>
<br>
Object b<br>
b set overrides {}<br>
b proc foo {obj} {<br>
puts "b: [self] foo"<br>
if {-1 == [lsearch [my set overrides] $obj]} {<br>
puts "b is overriding $obj foo"<br>
$obj forward foo [self] overridefoo $obj<br>
}<br>
}<br>
<br>
b proc overridefoo {obj args} {<br>
if {$args eq "done"} {<br>
puts "stop forwarding foo for $obj"<br>
set idx [lsearch -exact [my set overrides] $obj]<br>
my set overrides [lreplace [my set overrides] $idx $idx]<br>
# This doesn't work! What should be here?<br>
$obj forward foo {}<br>
puts "redispatching foo to native $obj foo"<br>
$obj foo $args<br>
} else {<br>
puts "using override of foo on $obj $args"<br>
}<br>
}<br>
a foo 123<br>
b foo a<br>
a foo 234<br>
a foo done <br>
<br>
----------<br>
<br>
>From what I gather, the intended mechanism for doing this is to use a mixin to intercept the foo method, but a already has a foo method that's being used for other purposes. I suppose I could dynamically create another object (or would it have to be a class?) to use as a mixin, but that seems unnecessarily complex. Is there a better way to accomplish what I'm trying to do? <br>
<br>
<br>
BTW, I'm running xotcl 1.3.6. If I run the above sample 3 times, I get the following error:<br>
<br>
procedure "foo" has formal parameter "::xotcl::initProcNS" that is not a simple name<br>
<br>
and the interpreter crashes on the next command. That's probably a bug, but since I'm not at the latest release, I don't know if it's something that's been fixed already.<br>
<br>
Thanks.<br>
<br>
Scott<br>
<br>
<br>
<br>
<br>
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