Python 局部变量行为异常

Question

这个问题已经有答案了:

Is it possible to modify a variable in python that is in an outer (enclosing), but not global, scope? (9 个回答)

已关闭 5 年前。

考虑以下代码:

def g():
    a = {}
    b = 0
    def f():
        a[0] = b
    f()
    print(a)
    return a
a = g()
print(a)

执行时会给出以下输出:

{0: 0}
{0: 0}

但是如果我尝试更新 b在f()如

def g():
    a = {}
    b = 0
    def f():
        a[0] = b
        b+=1
    f()
    print(a)
    return a
a = g()
print(a)

它抛出以下错误:

UnboundLocalError: local variable 'b' referenced before assignment

这是预期的吗？我需要更新b里面f() 。这不可能吗？

Answer 1

将 a 和 b 显式传递给您的函数。

def g():
    a = {}
    b = 0
    def f(a, b):
        a[0] = b
        b += 1
        return b
    b = f(a, b)
    return a