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在函数里嵌套函数，嵌套函数如果使用到了外面的变量，则使用的是外面函数栈中的变量，相当于是浅复制一份给内函数，直到外函数执行结束才会复制一份给内函数
def fun_list():
fun_list=[]
for i in range(1,10):
print('the id of i is:',id(i))
def fun():
'''
定义用到i，i是fun_list2()内存中的i，改变了i的值函数打印的也会变
'''
print('I have got number %d, '%i,'its id is:',id(i))
fun()
print('a new fun() is defined')
fun_list.append(fun)
'''
函数执行结束时，才会把i复制传递给之前定义的各个fun()
'''
return fun_list
fun_list=fun_list()
print('------------------------')
for i in range(len(fun_list)):
fun_list[i]()
运行结果
the id of i is: 1652060816 I have got number 1, its id is: 1652060816 a new fun() is defined the id of i is: 1652060848 I have got number 2, its id is: 1652060848 a new fun() is defined the id of i is: 1652060880 I have got number 3, its id is: 1652060880 a new fun() is defined the id of i is: 1652060912 I have got number 4, its id is: 1652060912 a new fun() is defined the id of i is: 1652060944 I have got number 5, its id is: 1652060944 a new fun() is defined the id of i is: 1652060976 I have got number 6, its id is: 1652060976 a new fun() is defined the id of i is: 1652061008 I have got number 7, its id is: 1652061008 a new fun() is defined the id of i is: 1652061040 I have got number 8, its id is: 1652061040 a new fun() is defined the id of i is: 1652061072 I have got number 9, its id is: 1652061072 a new fun() is defined ------------------------ I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 I have got number 9, its id is: 1652061072 Process finished with exit code 0
再举一个例子：
def func():
i=1
print('The number of i is: %d now'%i)
def i_printer():
print('Hello! The number of i is %d'%i)
for i in range(2,10):
print('The number of i is: %d now'%i)
i_printer()
return i_printer
func()
虽然i_printer()在for之前就定义了，但是此是它里面使用的i是func()的栈里面的i，i变了打印的内容就会变，我们运行一下
The number of i is: 1 now i_printer: The number of i is: 2 now Hello! The number of i is 2 i_printer: The number of i is: 3 now Hello! The number of i is 3 i_printer: The number of i is: 4 now Hello! The number of i is 4 i_printer: The number of i is: 5 now Hello! The number of i is 5 i_printer: The number of i is: 6 now Hello! The number of i is 6 i_printer: The number of i is: 7 now Hello! The number of i is 7 i_printer: The number of i is: 8 now Hello! The number of i is 8 i_printer: The number of i is: 9 now Hello! The number of i is 9
解决方案
def fun_list():
fun_list=[]
for i in range(1,10):
print('the id of i is:',id(i))
def new_fun(i):
'''
解决方案：在外面再嵌套一层函数，把内层函数用到的外层参数传给它，然后在里面定义内函数，把内层函数作为返回值，调用这个函数来定义内层函数
'''
def fun():
'''
定义用到i，i是fun_list2()内存中的i，改变了i的值函数打印的也会变
'''
print('I have got number %d, '%i,'its id is:',id(i))
return fun
print('a new fun() is defined')
fun_list.append(new_fun(i))
'''
new_fun执行结束时，必须把i深复制给它的内层函数fun()
'''
return fun_list
fun_list=fun_list()
print('------------------------')
for i in range(len(fun_list)):
fun_list[i]()
the id of i is: 1652060816 a new fun() is defined the id of i is: 1652060848 a new fun() is defined the id of i is: 1652060880 a new fun() is defined the id of i is: 1652060912 a new fun() is defined the id of i is: 1652060944 a new fun() is defined the id of i is: 1652060976 a new fun() is defined the id of i is: 1652061008 a new fun() is defined the id of i is: 1652061040 a new fun() is defined the id of i is: 1652061072 a new fun() is defined ------------------------ I have got number 1, its id is: 1652060816 I have got number 2, its id is: 1652060848 I have got number 3, its id is: 1652060880 I have got number 4, its id is: 1652060912 I have got number 5, its id is: 1652060944 I have got number 6, its id is: 1652060976 I have got number 7, its id is: 1652061008 I have got number 8, its id is: 1652061040 I have got number 9, its id is: 1652061072 Process finished with exit code 0
