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测试题:
0.
global
1.
nonlocal
2.
外部函数的外部环境不能访问外部函数的内部函数,只有外部函数能访问内部函数.
3.
因为内部函数以相同的变量名,给外部函数的变量赋值的时候,此刻,外部函数的变量会被屏蔽,内部函数的变量可以看做临时申请的局部变量(因为是堆栈形式,如果是列表或者容器中的元素就可以直接使用),所以在内部函数直接使用外部函数的同名变量,且不事先赋值,会出错.(不建议在内部函数给外部函数的变量从新赋值)
4.
funOut()
5.
funOut()()
5.
6
6
6
动动手:
0.
def count():
string = '''%%$@_$^__#)^)&!_+]!*@&^}@[@%]()%+$&[(_@%+%$*^@$^!+]!&_#)_*}{}}!}_]$[%}@[{_@#_^{*
@##&{#&{&)*%(]{{('''
#找出所有独立的字符串
alltype = str()
for each in string:
if each not in alltype:
alltype += each
#找出每个字符串出现的次数,并对应装在L里
L = list()
for each in alltype:
num = 0
for each1 in string:
if each1 == each:
num += 1
L.append(num)
#打印字数
for each in range(len(alltype)):
## print('字符',alltype[each],'共出现',L[each],'次')
print('字符 %s 出现 %d 次' % (alltype[each],L[each]))
count()
1.
def findpsw():
string = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc
'''
L = str()
for each in range(len(string)):
if string[each].isalpha() and string[each].islower() and string[each-1].isupper() and string[each-2].isupper() and string[each-3].isupper() and string[each + 1].isupper() and string[each + 2].isupper() and string[each+3].isupper():
L = L + string[each]
print(L)
findpsw()
2.课堂笔记:
1. 函数里想用全局变量,用global
2. 内嵌函数想用外部函数的参数,用nonlocal声明
3. f()指向ruturn的值,f本身是function,记录了函数变量的地址
4. 遇到bug,从特性解决,即找到一个特性点,验证是否可用,若特性点不可用,则整体不可用,若特性点可用,说明是某个其他点的问题
5. 第六题有些不懂,暂时不看 |
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