Variable scope for nested functions in python

Question

I have a demo example:

def outer(func):
 def inner(foo): 
 arg_inner=some_value 
 func(arg_inner,foo) 
 return inner
 def third_party(arg1,arg2):
 #do something
 final = outer(third_party)
 final(3)

I wonder how comes the inner function knows the value of func while being called. How the scope of inner function and outer are connected?

Answer 1

The compiler connects them.

When compiling outer and inner, the func name in outer is marked as a closure, and a free variable in inner. On execution, the interpreter then knows to attach a reference to the func name from outer to the inner function object as a closure cell.

You can see the result in the disassembly of the bytecode:

>>> import dis
>>> dis.dis(outer)
 2 0 LOAD_CLOSURE 0 (func)
 3 BUILD_TUPLE 1
 6 LOAD_CONST 1 (<code object inner at 0x1079a5ab0, file "<stdin>", line 2>)
 9 MAKE_CLOSURE 0
 12 STORE_FAST 1 (inner)
 4 15 LOAD_FAST 1 (inner)
 18 RETURN_VALUE 
>>> inner = outer(lambda a, b: None)
>>> dis.dis(inner)
 3 0 LOAD_DEREF 0 (func)
 3 LOAD_GLOBAL 0 (arg_inner)
 6 LOAD_FAST 0 (foo)
 9 CALL_FUNCTION 2
 12 POP_TOP 
 13 LOAD_CONST 0 (None)
 16 RETURN_VALUE 

The LOAD_CLOSURE wraps the func name in a closure for inner to use; MAKE_CLOSURE builds a function object (from the byte code object loaded with LOAD_CONST) with attached closure cells as a tuple. In inner, the LOAD_DEREF opcode loads the value from the func closure.

The closure can be found back on the resulting inner function object:

>>> inner.func_closure
(<cell at 0x107a25a28: function object at 0x107a28b18>,)
>>> inner.func_code.co_freevars
('func',)
>>> inner.func_closure[0].cell_contents
<function <lambda> at 0x107a28b18>

so the inner function object carries the closure with it for later dereferencing.

Answer 2

look at http://docs.python.org/2/reference/executionmodel.html#naming-and-binding.It contains explanations about scops.