On 8/07/2006 11:01 AM, tac-tics wrote: > Experimenting, I found that > >>>> x.fun = lambda: fun(x) > > works as well. Any comments on this way? >
Appears to work. Less typing for a no-argument method, but you need to
specify the argument list *twice* compared with *zero* times with the
types.MethodType way of doing it.
>>> x=K()
>>> x.metho = lambda(a): own(x, a)
>>> x.metho('abcde')
own
>>> x.ozz
'abcde'
>>>
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