Bind local variable to nested function in python

Asked Apr 26 '18 at 11:55

Active Aug 19 '22 at 12:39

Viewed 184 times

I'm trying to created nested functions in a loop in python, so each function would return a specific value depending on the value of said looped iterator.

Something like this:

for i in range(0, len(servers)):
    f = lambda: print(servers[i]['servername'])
    functions.append(f)

I would call the functions later on in the code an expect each function to return a given servername. But I always end up with all the functions returning the same thing (the last one) because they are all referring to the same thing, and not their local values.

Is what I'm trying to do at all possible? Or should I try this in another way?

asked Apr 26 '18 at 11:55

Christian Dechery

4

Use `lambda i=i: ...` to bind on definition. – Andras Deak -- Слава Україні Apr 26 '18 at 11:56
1

Ok, that worked! Also works with `def func(s=s)` which suits me even better. – Christian Dechery Apr 26 '18 at 12:38
Indeed. The point is that default keyword arguments are always set on definition. Whether it's a lambda or a proper function, it doesn't matter. – Andras Deak -- Слава Україні Apr 26 '18 at 12:39
This is actually the same reason why you need to watch out with [mutable default arguments](https://stackoverflow.com/questions/1132941/least-astonishment-and-the-mutable-default-argument). – Andras Deak -- Слава Україні Apr 26 '18 at 12:49

Bind local variable to nested function in python

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