sigurd at 12move.de
Fri Mar 19 20:19:52 CET 2004
On 19 Mar 2004, Darabos Daniel <- cyhawk at sch.bme.hu wrote:
>>>> def p( x ):
> ... print x
>>>> l = 
>>>> for i in range( 5 ):
> ... l.append( lambda: p( i ) )
>>>> for k in l:
> ... k()
> And it surprised me a little. I was expecting to see 0, 1, 2, 3, 4.
The `i' in your code is bound to the global value of `i' which is 4.
> After some brainwork I now kind of understand what happens and I even
> found a solution like this:
>>>> def mylambda( fn, *args ):
> ... return lambda: apply( fn, args )
> But I still feel a bit unsatisfied. Do you have some advice for me?
You have to bind the value of `i' in the lambda form; you did that with
your function definition but it can be done a bit simpler:
>>> funs = [lambda i=i: sys.stdout.write(str(i)+"\n") for i in range(4)]
>>> for f in funs:
With the `i=i' the `i' in the lambda got bound.
And has thou slain the Jabberwock?
Come to my arms, my beamish boy!
O frabjous day! Callooh! Callay!'
He chortled in his joy. "Lewis Carroll" "Jabberwocky"
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