在这个修改后的代码中,我们为local_var预先定义了一个默认值(None),从而确保它在所有情况下都会被定义。 总之,解决“cannot access local variable”的问题需要理解变量的作用域和生命周期,并根据具体情况选择合适的解决方案。
UnboundLocalError: cannot access local variable 'file' where it is not associated with a value 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. UnboundLocalError异常是NameError异常的子类,异常信息提示没有找到 file 变量,这是因为 open(filename) 6.2 else 代码块 与while 和 for 循环类型,try 语句也可以带...
问重新分配到更新变量时，在函数范围外访问python变量EN变量的范围是在其中可见变量的代码区域。变量作用域...
money += value UnboundLocalError: local variable 'money' referenced before assignment 可以看到,调用函数add_money后抛出错误UnboundLocalError.根据报错信息,局部变量(local variable)money在赋值前被引用。那么,什么是局部变量?这个错误是如何产生的? 1. 局部变量与全局变量 在Python中,如果变量名出现...
a = 1 print(id(a)) # 140724533279160 def bar(): print(locals()) a = a + 1 # (13) print(locals()) return a bar() ''' {} UnboundLocalError: cannot access local variable 'a' where it is not associated with a value ''' 示例三。在函数内部,通过global 关键词,声明变量名 a 是来...
inhahe changed the title incorrectly reporting "cannot access local variable ... where it is not associated with a value" when using += reporting "cannot access local variable ... where it is not associated with a value" in the wrong place when using += Nov 26, 2024 Member skirpichev...
# Local var x not the same as global variable x x = num # => 43 print(x) # => 43 def set_global_x(num): global x print(x) # => 5 x = num # global var x is now set to 6 print(x) # => 6 set_x(43) set_global_x(6) ...
local scope will change global variable due to same memory used input: importnumpyasnpdeftest(a):a[0]=np.nanm=[1,2,3]test(m)print(m) output: [nan, 2, 3] Note python has this really weird error if you define local variable in a function same name as the global variable, program...
Because this will define the variable inside the function's scope. It will no longer go to the surrounding (global) scope to look up the variables value but will create a local variable that stores the value of x at that point in time.funcs = [] for x in range(7): def some_func(...
But when trying to access and then print the same variable from outside the function (# Print statement 2), it raised a NameError. This is because first_num is "local" to the function - thus, it cannot be reached from outside the function body. Enclosing Scope What if we have a ...