问题
here is my code:
def f(x):
def g(n):
if n < 10:
x = x + 1
g(n + 1)
g(0)
When I evaluate f(0), there would be an error "x referenced before assignment".
However, when I use "print x" instead of "x = x + 1" , it will work.
It seems that in the scope of g, I can only use x as an "use occurrence" but not a "binding occurrence". I guess the problem is that f passes to g only the VALUE of x.
Am I understanding it correctly or not? If not, can someone explain why the left side of "x = x + 1" is not defined before reference?
Thanks
回答1:
You are understanding it correctly. You cannot use x
to assign to in a nested scope in Python 2.
In Python 3, you can still use it as a binding occurrence by marking the variable as nonlocal
; this is a keyword introduced for just this usecase:
def f(x):
def g(n):
nonlocal x
if n < 10:
x = x + 1
g(n + 1)
g(0)
In python 2, you have a few work-arounds; using a mutable to avoid needing to bind it, or (ab)using a function property:
def f(x):
x = [x] # lists are mutable
def g(n):
if n < 10:
x[0] = x[0] + 1 # not assigning, but mutating (x.__setitem__(0, newvalue))
g(n + 1)
g(0)
or
def f(x):
def g(n):
if n < 10:
g.x = g.x + 1
g(n + 1)
g.x = x # attribute on the function!
g(0)
回答2:
Yes, assigning to names is different than reading their values. Any names that are assigned to in a function are considered local variables of that function unless you specify otherwise.
In Python 2, the only way to "specify otherwise" is to use a global
statement to allow you assign to a global variable. In Python 3, you also have the nonlocal
statement to assign a value to a variable in a higher (but not necessarily global) scope.
In your example, x
is in a higher but non-global scope (the function f
). So there is no way to assign to x
from within g
in Python 2. In Python 3 you could do it with a nonlocal x
statement in g
.
来源:https://stackoverflow.com/questions/14678434/passing-argument-from-parent-function-to-nested-function-python