Python - locals（）和闭包

Question

I can not find an adequate explanation for this behavior.

>>> def a():
...     foo = 0
...     print locals()
...     def b():
...         print locals()
...     b()

>>> a()
{'foo': 0}
{}

But:

>>> def a():
...     foo = 0
...     print locals()
...     def b():
            foo
...         print locals()
...     b()

>>> a()
{'foo': 0}
{'foo': 0}

I understand that in the second case there is a closure, but I can not find a detailed description of what actually is and under what conditions should return the function locals() .

Answer 1

If you don't assign to foo within the closure, Python resolves it to the foo of the scope one level up (and on up until it finds a foo somewhere or throws an exception).

By mentioning foo within b() in the second example, you put foo into the locals within b() , but it resolves to the foo within the body of a() . If you assign, say, foo = 1 in b() , you would see

 {'foo': 0}
 {'foo': 1}

as the output.

Answer 2

locals() built-in function prints local symbol table which is bound to a code object , and filled up when interpreter receives a name in a source code.

Second example, when disassembled, will contain LOAD_GLOBAL foo bytecode instruction in b function code. This LOAD_GLOBAL instruction will move up the scope, find outer foo name and bind it to the code object by adding name offset into co_names attribute of the closure's (function b) code object.

locals() function prints local symbol table (as was said before, co_names attribute of function's code object).

Read more about code objects here .