Modify bound variables of a closure in Python

Question

Is there any way to modify the bound value of one of the variables inside a closure? Look at the example to understand it better.

def foo():
 var_a = 2
 var_b = 3

 def _closure(x):
  return var_a + var_b + x

 return _closure


localClosure = foo()

# Local closure is now "return 2 + 3 + x"
a = localClosure(1) # 2 + 3 + 1 == 6

# DO SOME MAGIC HERE TO TURN "var_a" of the closure into 0
# ...but what magic? Is this even possible?

# Local closure is now "return 0 + 3 + x"
b = localClosure(1) # 0 + 3 +1 == 4

Answer 1

Why not make var_a and var_b arguments of the function foo?

def foo(var_a = 2, var_b = 3):
    def _closure(x):
        return var_a + var_b + x
    return _closure

localClosure = foo() # uses default arguments 2, 3
print localClosure(1) # 2 + 3 + 1 = 6

localClosure = foo(0, 3)
print localClosure(1) # 0 + 3 + 1 = 4

Answer 2

I don't think there is any way to do that in Python. When the closure is defined, the current state of variables in the enclosing scope is captured and no longer has a directly referenceable name (from outside the closure). If you were to call foo() again, the new closure would have a different set of variables from the enclosing scope.

In your simple example, you might be better off using a class:

class foo:
        def __init__(self):
                self.var_a = 2
                self.var_b = 3

        def __call__(self, x):
                return self.var_a + self.var_b + x

localClosure = foo()

# Local closure is now "return 2 + 3 + x"
a = localClosure(1) # 2 + 3 + 1 == 6

# DO SOME MAGIC HERE TO TURN "var_a" of the closure into 0
# ...but what magic? Is this even possible?
localClosure.var_a = 0

# Local closure is now "return 0 + 3 + x"
b = localClosure(1) # 0 + 3 +1 == 4

If you do use this technique I would no longer use the name localClosure because it is no longer actually a closure. However, it works the same as one.

Answer 3

I've found an alternate answer answer to Greg's, slightly less verbose because it uses Python 2.1's custom function attributes (which conveniently enough can be accessed from inside their own function).

def foo():
 var_b = 3

 def _closure(x):
  return _closure.var_a + var_b + x

 _closure.func_dict['var_a'] = 2
 return _closure


localClosure = foo()

# Local closure is now "return 2 + 3 + x"
a = localClosure(1) # 2 + 3 + 1 == 6

# DO SOME MAGIC HERE TO TURN "var_a" of the closure into 0
# ...but what magic? Is this even possible?
# apparently, it is
localClosure.var_a = 0

# Local closure is now "return 0 + 3 + x"
b = localClosure(1) # 0 + 3 +1 == 4

Thought I'd post it for completeness. Cheers anyways.

Answer 4

It is quite possible in python 3 thanks to the magic of nonlocal.

def foo():
        var_a = 2
        var_b = 3

        def _closure(x, magic = None):
                nonlocal var_a
                if magic is not None:
                        var_a = magic

                return var_a + var_b + x

        return _closure


localClosure = foo()

# Local closure is now "return 2 + 3 + x"
a = localClosure(1) # 2 + 3 + 1 == 6
print(a)

# DO SOME MAGIC HERE TO TURN "var_a" of the closure into 0
localClosure(0, 0)

# Local closure is now "return 0 + 3 + x"
b = localClosure(1) # 0 + 3 +1 == 4
print(b)

Answer 5

We've done the following. I think it's simpler than other solutions here.

class State:
    pass

def foo():
    st = State()
    st.var_a = 2
    st.var_b = 3

    def _closure(x):
        return st.var_a + st.var_b + x
    def _set_a(a):
        st.var_a = a

    return _closure, _set_a


localClosure, localSetA = foo()

# Local closure is now "return 2 + 3 + x"
a = localClosure(1) # 2 + 3 + 1 == 6

# DO SOME MAGIC HERE TO TURN "var_a" of the closure into 0
localSetA(0)

# Local closure is now "return 0 + 3 + x"
b = localClosure(1) # 0 + 3 +1 == 4

print a, b