Why static binding works differently for class and function?

问题

In python (tested on 2.7.6) all variables are statically bound to a scope at compile time. This process is well described in http://www.python.org/dev/peps/pep-0227/ and http://docs.python.org/2.7/reference/executionmodel.html

It is explicitly stated that "If a name binding operation occurs anywhere within a code block, all uses of the name within the block are treated as references to the current block."

A function is a code block so the following code with fail because x is assigned after its use (so at compile time it is defined local because it is assigned somewhere in the function, but at execution time, it is used before being bound).

x = 1
def f():
    print x 
    x = 2
    print x

>>> f()

Traceback (most recent call last):
  File "<pyshell#46>", line 1, in <module>
    f()
  File "<pyshell#45>", line 2, in f
    print x
UnboundLocalError: local variable 'x' referenced before assignment

A class is also a code block, so we should observe exactly the same behavior. But this is not what I observe. Look at this example:

x = 1
class C():
    y = x + 10
    x = 2
    def __init__(self):
        print C.y

>>> C.x
2
>>> C.y
11      
>>> C()
11
<__main__.C instance at 0x00000000027CC9C8>

As the class definition is a code block, any assignment within this block should make the variable local. So x should be local to the class C, so y = x + 10 should result in an UnboundLocalError. Why there is not such error?

回答1:

Yes - it seems that the documentation is rather misleading. A class definition doesn't actually work quite the same as other normal blocks:

global_one = 0

class A(object):
    x = global_one + 10
    global_one = 100
    y = global_one + 20
    del global_one
    z = global_one + 30

a = A()
print a.x, a.y, a.z, global_one

results in: 10, 120, 30, 0

if you try the same thing with a function, you get an UnboundLocalError on your first access of global_one.

The reason for this is that class definitions as normal have access to the parent scope, however, all name assignments do NOT modify a local scope, but in fact are captured into the class's data attributes dictionary. There are hints about this in the documentation, but it's certainly not obvious.

来源：https://stackoverflow.com/questions/20148256/why-static-binding-works-differently-for-class-and-function

标签

python

class

scope

static-binding