代码不报错不给结果是什么情况？
Python课后作业代码，写好后运行不给结果，也不报错。请问是什么情况？运行结果如图，代码如下
def find1(str1):
a = [] #建立空列表储存转化为大小写判断的flag
b = '' #建议空字符串以便拼接密码
for each in str1:
if each.islower() == 1: #判断是否为小写
flag = 1 #将小写标记为1
a.append(flag) #将flag储存到列表a中
else:
flag = 0 #将大写标记为0
a.append(flag) #将flag储存到列表a中
i = 3 #按照题目要求，从第4个元素开始判断
while i <= len(a)-3: #在倒数第4个元素处停止
if i == 3 or i == (len(a)-3): #在第4个和倒数第4个处判断的组合
if a == 1 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0:
b = b + str1
else: #在其他位置上的元素判断的组合
if a == 1 and a == 1 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 1:
b = b + str1
print('密码是' + c)
x = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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find1(x)
比较新奇的算法了，就是效率有点低，不过对于现在的电脑来说不算什么，先说两个显性错误：
1.while循环中i值没有变化，死循环了，加一条i+=1或者换成for循环；
2.变量c没有，是变量b。
然后是算法bug：
1.没有考虑非字母，这里主要是'\n'；
2.判断第四个字符和倒数第四个字符不是简单看前三后三，而是第四个是前三后四，倒数第四个是前四后三。
修正了代码，可以跑起来了，还有可以优化的地方再慢慢想吧：
def find1(str1):
a = [] #建立空列表储存转化为大小写判断的flag
b = '' #建议空字符串以便拼接密码
for each in str1:
if each.islower() == 1: #判断是否为小写
flag = 1 #将小写标记为1
a.append(flag) #将flag储存到列表a中
elif each.isupper() == 1:
flag = 0 #将大写标记为0
a.append(flag) #将flag储存到列表a中
else:
flag = 2 #非字母标记为2
a.append(flag) #将flag储存到列表a中
i = 3 #按照题目要求，从第4个元素开始判断
while i <= len(a)-3: #在倒数第4个元素处停止
if i == 3 : #在第4个处判断的组合
if a == 1 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a != 0:
b = b + str1
if i == (len(a)-3): #在倒数第4个处判断[非0，0，0，0，1，0，0，0]的组合
if a == 1 and a != 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0:
b = b + str1
else: #在其他位置上的元素判断[非0，0，0，0，1，0，0，0，非0]的组合
if a == 1 and a != 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a != 0:
b = b + str1
i+=1
print('密码是' + b) 还在算吧 本帖最后由 heidern0612 于 2020-5-13 07:39 编辑
大致看了下，最后while的时候，i的变化呢？
i什么时候不满足while呢？
没退出的话，应该一直就循环，卡在循环里了吧？
你可以在if分支下面建个print，打印一句话。然后运行下，你会发现一直在打印这句话。
也就是说，你i没有小于你while条件的时候，自然就一直循环了。
你可以运行我这段代码试试，我就加了俩print,别的代码没动你的。
def find1(str1):
a = [] #建立空列表储存转化为大小写判断的flag
b = '' #建议空字符串以便拼接密码
for each in str1:
if each.islower() == 1: #判断是否为小写
flag = 1 #将小写标记为1
a.append(flag) #将flag储存到列表a中
else:
flag = 0 #将大写标记为0
a.append(flag) #将flag储存到列表a中
i = 3 #按照题目要求，从第4个元素开始判断
while i <= len(a)-3: #在倒数第4个元素处停止
if i == 3 or i == (len(a)-3): #在第4个和倒数第4个处判断的组合
if a == 1 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0:
b = b + str1
print("卡在if")
else: #在其他位置上的元素判断的组合
if a == 1 and a == 1 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 0 and a == 1:
b = b + str1
print("卡在else")
print('密码是' + c)
x = '''ACFlCTLIQlAIVMTqHFkswqbDDHtpgcWaXSSglUYKE
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qKWGc'''
find1(x)
对的，没有出现下一个“>>>”应该是还在算 {:10_256:}{:10_256:}刷个存在感 heidern0612 发表于 2020-5-13 07:37
大致看了下，最后while的时候，i的变化呢？
i什么时候不满足while呢？
感谢回复，确实卡在了if的位置。我都不知道啥时候不小心把i++给删掉了。我还一直没注意到。 txxcat 发表于 2020-5-12 23:56
比较新奇的算法了，就是效率有点低，不过对于现在的电脑来说不算什么，先说两个显性错误：
1.while循环中i ...
感谢老哥，前面写了i += 1后面不小心删掉没注意到。最后的b和c也是后面整理代码出现的错误。算法bug也指出来了。太棒了！{:10_275:}{:10_275:}
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