newbie: binding args in callbacks

[Edward K. Ream]

> My suggestion:
> 
>     def makeCallback(self, val):
>         def callback(self=self, val=val): return self.myCallback(val)
>         return callback
> 
> and then:
> 
>     b = Tk.Button(..., command=self.makeCallback(val) )
> 
> You don't need the default-values trick in Python 2.2 -- so, if
> that's what you're using, it would be better to use, instead:
> 
>     def makeCallback(self, val):
>         def callback(): return self.myCallback(val)
>         return callback

I tried this, and indeed this does work on 2.2. Really.  Thanks very
much.

> If you're insistent on using lambda (only sensible reason being that
> you made a bet about lambda being used here), you _can_, e.g. in 2.2:
> 
>     def makeCallback(self, val):
>         return lambda: self.myCallback(val)
> 
> but I think the nested-function approach is more readable.

This also works on 2.2.

This "extra" level of indirection is needed. The following does not
work:

for val in vals:
  callback=lambda: self.myCallback(val) # or (val=val)
   b = Tk.Button(..., command=callback)

All the callbacks get bound to a function with the last val in vals.  I
have no idea why.  Maybe you could explain?

BTW, the reason I really like Python is that someone like me with
limited knowledge can actually produce some real work.  Anyway, thanks
again for your help!

Edward
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Edward K. Ream   email:  edream at tds.net
Leo: Literate Editor with Outlines
Leo: http://personalpages.tds.net/~edream/front.html
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