UnboundLocalError: cannot access local variable 'count' where it is not associated with a value 这是为啥呢？在 increment 函数内部，count += 1 这行代码，Python 默认会把 count 当成 increment 函数内部的局部变量。 但是，count 并没有在 increment 函数内部定义，所以就报错了。三、nonlocal关键字：打破...
AI代码解释 defouter():x=10print(f"original variable: x = {x} (id: {id(x)})")definner():nonlocal x x=20print(f"Inner function: x = {x} (id: {id(x)})")inner()print(f"Outer function: x = {x} (id: {id(x)})")outer() 输出 代码语言:javascript 代码运行次数:0 运行 AI...
here x is a local variable, it is visible just in the function of fun(). It has no relationship with the x variable outside. ''' x=10 deffun(): x=10000 print(x) fun() print(x) # In[] ''' result: UnboundLocalError: local variable 'x' referenced before assignment you can do ...
Nonlocal variable must be bound in an outer function scope. 代码语言:javascript 代码运行次数:0 运行 AI代码解释 def make_counter(): count = 0 def counter(): nonlocal count count += 1 return count return counter test = make_counter() print(test()) print(test()) 运行结果: 代码语言:...
python x = "global variable" def outer_function(): x = "local to outer" def inner_function_global(): global x x = "modified by global in inner" print(x) def inner_function_nonlocal(): nonlocal x x = "modified by nonlocal in inner" print(x) inner_function_global() # 修改全局...
结果:UnboundLocalError: local variable 'count' referenced before assignment 原因:原因跟上面 实例1 一样只不过不是全局变量而是 内层函数 修改外部函数前没有事先声明 python解释器在内层函数里找不到 count 变量就会直接报错 实例2: Copy deffunc1(): ...
)>>>f()梯阅线条# 不可修改嵌套作用域变量值>>>defoutf():a='梯阅线条'definf():print(a)a='tyxt'returninf>>>f=outf()>>>f()Traceback (mostrecentcalllast):File"<pyshell#28>", line1, in<module>f()File"<pyshell#26>", line4, ininfprint(a)UnboundLocalError: localvariable'a'...
实例(Python 3.0+)#!/usr/bin/python3 a = 10 def test(): a = a + 1 print(a) test()以上程序执行,报错信息如下:Traceback (most recent call last): File "test.py", line 7, in <module> test() File "test.py", line 5, in test a = a + 1UnboundLocalError: local variable 'a' ...
UnboundLocalError: cannot access local variable 'prev' where it is not associated with a value解释:nonlocal 是Python中的关键字,用于声明一个嵌套函数中的变量是来自于其外部函数的局部作用域,而不是来自于全局作用域。 在这段代码中,prev 和head 是在recursion 函数之外定义的变量。然而,我们需要在 recursion...
(2)nonlocal的作用范围仅对于所在子函数的上一层函数中拥有的局部变量,必须在上层函数中已经定义过,并且不是非全局变量,否则报错。Nonlocal variable must be bound in an outer function scope. defmake_counter(): count =0defcounter():nonlocalcount ...