您可以分配给父函数中定义的变量吗? [英] Can you assign to a variable defined in a parent function?
问题描述
可能的重复:
Python 嵌套函数变量作用域
经过多次反复试验,我最终发现这行不通:
After much trial and error I have eventually discovered that this doesn't work:
def a():
def b():
print x
x=2
x = 1
b()
print x
您得到一个异常(x 在被引用之前未定义).所以看起来 b 可以从 x 读取,但是如果它试图赋值给它,Python 会将它对 'x' 的解释更改为一个局部变量,现在没有定义.
You get an exception (x not defined before being referenced). So it looks like b can read from x, but if it tries to assign to it, Python changes its interpretation of 'x' to be a local variable, which is now not defined.
问我自己病态的好奇心:有没有办法实现这一点?有没有办法显式访问父函数的范围?(x 不是全局的)
Question for my own sick curiosity: is there any way of achieving this? Is there a way of explicitly accessing the scope of the parent function? (x is not global)
推荐答案
nonlocal
语句 可以做到这一点.
The nonlocal
statement in Python 3 will do this.
编辑:在 Python 2 中,没有一种简单的方法可以做到.如果您需要此功能,我建议您使用一些可变容器对象.例如:
Edit: In Python 2, there's not a simple way to do it. I suggest you use some mutable container object if you need this capability. For example:
def a():
def b():
print d["x"]
d["x"]=2
d = dict(x=1)
b()
print d["x"]
如果您绝对必须为 CPython 2 模拟 nonlocal
,您可以通过 Python C API 以这种方式破解它:
If you absolutely must emulate nonlocal
for CPython 2, you can hack it with the Python C API this way:
import ctypes
import inspect
locals_to_fast = ctypes.pythonapi.PyFrame_LocalsToFast
locals_to_fast.restype = None
locals_to_fast.argtypes = [ctypes.py_object, ctypes.c_int]
def set_in_frame(frame, name, value):
frame.f_locals[name] = value
locals_to_fast(frame, 1)
def a():
def b(frame=inspect.currentframe()):
print x
set_in_frame(frame, "x", 2)
x = 1
b()
print x
您也可以将框架设置为本地,而不是调用 PyFrame_LocalsToFast()
,您可以操作 a
的字节码,使其使用 LOAD_NAME
而不是 LOAD_FAST
.请不要做这些事情.对于您的用例,肯定有更好的解决方案.
You could also set the frame local, and instead of calling PyFrame_LocalsToFast()
, you could manipulate the bytecode of a
so that it uses LOAD_NAME
instead of LOAD_FAST
. Please don't do either of these things. There is surely a better solution for your use case.
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