我对python很新，所以请耐心等待。我正在组合一个小的加密拼图，其中一部分需要替换密码。在这种情况下，每个字母用三个随机字母替换。每次运行此代码时，当我尝试加密某些内容时都会抛出错误：
Traceback (most recent call last):
File "/home/pi/Desktop/expansion cipher.py", line 159, in <module>
crypt()
File "/home/pi/Desktop/expansion cipher.py", line 142, in crypt
print encrypt(txt)
File "/home/pi/Desktop/expansion cipher.py", line 127, in encrypt
ctxt = ctxt + exp_ciph(ptxt[counter])
File "/home/pi/Desktop/expansion cipher.py", line 121, in exp_ciph
return cur_exp
UnboundLocalError: local variable 'cur_exp' referenced before assignment
这是我的代码：
def rev_ciph(char):
if char == "gps":
cur_rev = "_"
if char == "evl":
cur_rev = "."
if char == "jkb":
cur_rev = "e"
if char == "bhj":
cur_rev = "t"
if char == "szk":
cur_rev = "a"
if char == "nwu":
cur_rev = "o"
if char == "dpl":
cur_rev = "i"
if char == "sbg":
cur_rev = "n"
if char == "dsl":
cur_rev = "s"
if char == "yhq":
cur_rev = "p"
if char == "sav":
cur_rev = "h"
if char == "gfs":
cur_rev = ","
if char == "rtg":
cur_rev = "d"
if char == "fqu":
cur_rev = "l"
if char == "rjt":
cur_rev = "u"
if char == "sbv":
cur_rev = "c"
if char == "yqm":
cur_rev = "m"
if char == "ywh":
cur_rev = "f"
if char == "drt":
cur_rev = "y"
if char == "zfd":
cur_rev = "w"
if char == "asn":
cur_rev = "g"
if char == "fzj":
cur_rev = "p"
if char == "rka":
cur_rev = "b"
if char == "kzv":
cur_rev = "v"
if char == "lah":
cur_rev = "k"
if char == "rma":
cur_rev = "x"
if char == "fqn":
cur_rev = "q"
if char == "vrq":
cur_rev = "j"
if char == "tfv":
cur_rev = "z"
return cur_rev
def exp_ciph(char):
if char == "_":
cur_exp = "gps"
if char == ".":
cur_exp = "evl"
if char == "e":
cur_exp = "jkb"
if char == "t":
cur_exp = "bhj"
if char == "a":
cur_exp = "szk"
if char == "o":
cur_exp = "nwu"
if char == "i":
cur_exp = "dpl"
if char == "n":
cur_exp = "sbg"
if char == "s":
cur_exp = "dsl"
if char == "p":
cur_exp = "yhq"
if char == "h":
cur_exp = "sav"
if char == ",":
cur_exp = "gfs"
if char == "d":
cur_exp = "rtg"
if char == "l":
cur_exp = "fqu"
if char == "u":
cur_exp = "rjt"
if char == "c":
cur_exp = "sbv"
if char == "m":
cur_exp = "yqm"
if char == "f":
cur_exp = "ywh"
if char == "y":
cur_exp = "drt"
if char == "w":
cur_exp = "zfd"
if char == "g":
cur_exp = "asn"
if char == "p":
cur_exp = "fzj"
if char == "b":
cur_exp = "rka"
if char == "v":
cur_exp = "kzv"
if char == "k":
cur_exp = "lah"
if char == "x":
cur_exp = "rma"
if char == "q":
cur_exp = "fqn"
if char == "j":
cur_exp = "vrq"
if char == "z":
cur_exp = "tfv"
return cur_exp
def encrypt(ptxt):
ctxt = "Ciphertext: "
counter = 0
while counter <= len(ptxt):
ctxt = ctxt + exp_ciph(ptxt[counter])
counter += 1
return ctxt
def decrypt(ctxt):
ptxt = "Plaintext: "
counter = 0
while counter <= len(ctxt):
ptxt = ptxt + rev_ciph(ctxt[counter])
counter += 1
return ptxt
def crypt():
print
txt = raw_input("Plaintext: ")
print encrypt(txt)
print
def ucrypt():
print
txt = raw_input("Ciphertext: ")
print decrypt(txt)
print
ex_code = False
while ex_code == False:
print "(1) Encrypt"
print "(2) Decript"
print "(3) Exit"
print
mchoc = raw_input("What would you like to do(1,2,3?): ")
if mchoc == "1":
crypt()
if mchoc == "2":
ucrypt()
if mchoc == "3":
ex_code = True
print
答案 0 :(得分：0)
你有两个问题。首先，您不能处理所有可能的输入。刚刚复制并测试了您的代码后，我发现只有当您尝试加密包含UnboundLocalError: local variable 'cur_exp' referenced before assignment（或exp_ciph(char)未处理的字符的内容时才会获得rev_ciph(char) if } 对于这个问题）。例如，这些函数都不处理大写字母。 Python总是区分大小写，并将大写和小写字母视为完全不同的字符。您的功能也不会处理其他字符，例如＆＃34;，＆＃34;和＆＃34;;＆＃34;。如果为程序提供一个要加密的字符串，其中包含前两个函数未处理的字符，那么，当该字符出现时，这些函数中cur_exp语句的所有条件都为false，因此{永远不会创建{1}}并为其分配值。当其中一个函数尝试return cur_exp时，它会发现cur_exp不存在。这是您在上面描述的错误的来源。
如果你输入的字符串只包含小写字母和可能的＆＃34; _＆＃34;，你会得到一个完全不同的错误，即：
IndexError: string index out of range
这是因为函数encrypt(ptxt)总是寻找一个字符而不是字符串。在Python（以及我认为，每种编程语言）中，可迭代中的元素（如字符串或列表）由OFFSET从第一个项目索引;也就是说，前面有多少项目。所以，如果你有一个包含4个字符的字符串foo：
foo = "abcd"
然后：
foo[0] == "a"
foo[1] == "b"
foo[2] == "c"
foo[3] == "d"
没有foo[4]。调用foo[4]将引发上述错误。 len(object)函数返回iterable中的项数。所以，按照上面的例子，
len(foo) == 4
然后调用
foo[len(foo)]
将提升IndexError:。这完全是encrypt(ptxt)函数
counter = 0
while counter <= len(ptxt):
ctxt = ctxt + exp_ciph(ptxt[counter])
counter += 1
您需要这样做：
counter = 0
while counter <= len(ptxt) - 1:
ctxt = ctxt + exp_ciph(ptxt[counter])
counter += 1
这是我遇到的第一个遇到的问题。我不能保证不会再有了。希望这会让你走得更远。