【ZZ已解決】Python中如何在嵌套函數內部訪問被嵌套（的父級函數）中的（局部，非全局）變量

以前就遇到這個問題了：

#!/usr/bin/python
# -*- coding: utf-8 -*-
"""
Function:
【已解決】Python中如何在嵌套函數內部訪問被嵌套（的父級函數）中的（局部，非全局）變量
 
http://www.crifan.com/python_access_parent_nesting_function_local_variable_from_nested_function
 
Author:     Crifan Li
Version:    2012-12-25
Contact:    admin at crifan dot com
"""
 
def accessVarFromNestedFunc():
    localVarInParent = 1;
     
    def nestedFunc(): #
        localVarInParent = localVarInParent + 1 ;
        print "In nested func, localVarInParent=",localVarInParent;
     
    nestedFunc();
    print "In current parent nesting func, localVarInParent=",localVarInParent;
     
if __name__ == "__main__":
    accessVarFromNestedFunc();

即，在嵌套函數內部，操做，父級函數中的變量，可是對應的父級函數中該變量，只是個普通的局部變量。

可是結果會出錯的：

D:\tmp\tmp_dev_root\python\access_var_from_nested_func>access_var_from_nested_func.py

Traceback (most recent call last):

  File "D:\tmp\tmp_dev_root\python\access_var_from_nested_func\access_var_from_nested_func.py", line 24, in <module>

    accessVarFromNestedFunc();

  File "D:\tmp\tmp_dev_root\python\access_var_from_nested_func\access_var_from_nested_func.py", line 20, in accessVarFromNestedFunc

    nestedFunc();

  File "D:\tmp\tmp_dev_root\python\access_var_from_nested_func\access_var_from_nested_func.py", line 17, in nestedFunc

    localVarInParent = localVarInParent + 1 ;

UnboundLocalError: local variable ‘localVarInParent’ referenced before assignment

可是一直也嘗試過，寫成global的形式，可是仍是無法解決。

惟一可行的是，須要定義一個全局的變量才能夠：

#!/usr/bin/python
# -*- coding: utf-8 -*-
"""
Function:
【已解決】Python中如何在嵌套函數內部訪問被嵌套（的父級函數）中的（局部，非全局）變量
 
http://www.crifan.com/python_access_parent_nesting_function_local_variable_from_nested_function
 
Author:     Crifan Li
Version:    2012-12-25
Contact:    admin at crifan dot com
"""
 
localVarInParent = 1;
 
def accessVarFromNestedFunc():
 
    def nestedFunc(): #
        global localVarInParent;
        localVarInParent = localVarInParent + 1 ;
        print "In nested func, localVarInParent=",localVarInParent;
     
    nestedFunc();
    print "In current parent nesting func, localVarInParent=",localVarInParent;
     
if __name__ == "__main__":
    accessVarFromNestedFunc();

可是很明顯，不是想要的效果。

不想要增長全局的變量，只想操做父級函數內的變量。





【解決過程】

1.後來終於看到一個正確的解釋了：

How do I change nesting function’s variable in the nested function

針對Python 2.x，則能夠改成：

#!/usr/bin/python
# -*- coding: utf-8 -*-
"""
Function:
【已解決】Python中如何在嵌套函數內部訪問被嵌套（的父級函數）中的（局部，非全局）變量
 
http://www.crifan.com/python_access_parent_nesting_function_local_variable_from_nested_function
 
Author:     Crifan Li
Version:    2012-12-25
Contact:    admin at crifan dot com
"""
 
def accessVarFromNestedFunc():
    localVarInParent = [1]; #here just define a list, first value is what we want to use
     
    def nestedFunc(): #
        localVarInParent[0] = localVarInParent[0] + 1 ; # localVarInParent[0] is the first value of above list value: localVarInParent, and its initial value is 1
        print "In nested func, localVarInParent[0]=",localVarInParent[0];#2,3,4,5,6
     
    for i in range(5):
        nestedFunc();
    # here can got value is 6, which is changed after nested function
    print "In current parent nesting func, localVarInParent[0]=",localVarInParent[0]; #In current parent nesting func, localVarInParent[0]= 6
     
if __name__ == "__main__":
    accessVarFromNestedFunc();

針對Python 3.x，是能夠添加對應的nonlocal的聲明的。這個暫時不去折騰了，有空再試試。 【總結】 Python中，嵌套函數內部去操做被嵌套的父級函數中的變量的話： Python 2.x：把變量弄進一個列表中的第1個值，index=0，而後就能夠在嵌套函數中，得到該list列表變量，操做其中第1個值了。 Python 3.x：把變量定義爲nonlocal便可。