为什么这个替换密码脚本不断抛出错误？
Why does this substitution cipher script keep throwing an error?
我是 python 的新手，请多多包涵。我正在拼凑一个小密码谜题，其中一部分需要替换密码。在这种情况下，每个字母都被三个随机字母替换。每次我 运行 此代码在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
你有两个问题。首先是您没有处理所有可能的输入。刚刚复制并测试了您的代码后，我看到当且仅当您尝试加密包含 exp_ciph(char)（或 rev_ciph(char)那件事）。例如，这些函数都不处理大写字母。 Python 始终区分大小写，并将大写字母和小写字母视为完全不同的字符。您的函数也不处理其他字符，例如“,”和“;”。如果您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 if 语句的所有条件都为假，因此 cur_exp 永远不会被创建和赋值。当其中一个函数尝试 return cur_exp 时，它发现 cur_exp 不存在。这就是您上面描述的错误的来源。
如果您输入的字符串仅由小写字母和可能的“_”组成，则会出现完全不同的错误，即：
IndexError: string index out of range
这是因为函数 encrypt(ptxt) 总是寻找比字符串多一个字符。在 Python（而且，我认为，每一种编程语言）中，可迭代对象（例如字符串或列表）中的元素是通过它们从第一项开始的 OFFSET 进行索引的；也就是说，它前面有多少项。因此，如果您有一个包含 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] 将恰好引发上述错误。 len(object) 函数 returns 迭代中的项目数。因此，与上面的示例保持一致，
len(foo) == 4
因此调用
foo[len(foo)]
将加注 IndexError:。这正是您的 encrypt(ptxt) 函数中经常发生的事情
counter = 0
while counter <= len(ptxt):
ctxt = ctxt + exp_ciph(ptxt[counter])
counter += 1
你会想要这样做：
counter = 0
while counter <= len(ptxt) - 1:
ctxt = ctxt + exp_ciph(ptxt[counter])
counter += 1
这是我遇到的前两个问题运行。我不保证没有更多。希望这能让你走得更远。
我是 python 的新手，请多多包涵。我正在拼凑一个小密码谜题，其中一部分需要替换密码。在这种情况下，每个字母都被三个随机字母替换。每次我 运行 此代码在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
你有两个问题。首先是您没有处理所有可能的输入。刚刚复制并测试了您的代码后，我看到当且仅当您尝试加密包含 exp_ciph(char)（或 rev_ciph(char)那件事）。例如，这些函数都不处理大写字母。 Python 始终区分大小写，并将大写字母和小写字母视为完全不同的字符。您的函数也不处理其他字符，例如“,”和“;”。如果您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 if 语句的所有条件都为假，因此 cur_exp 永远不会被创建和赋值。当其中一个函数尝试 return cur_exp 时，它发现 cur_exp 不存在。这就是您上面描述的错误的来源。
如果您输入的字符串仅由小写字母和可能的“_”组成，则会出现完全不同的错误，即：
IndexError: string index out of range
这是因为函数 encrypt(ptxt) 总是寻找比字符串多一个字符。在 Python（而且，我认为，每一种编程语言）中，可迭代对象（例如字符串或列表）中的元素是通过它们从第一项开始的 OFFSET 进行索引的；也就是说，它前面有多少项。因此，如果您有一个包含 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] 将恰好引发上述错误。 len(object) 函数 returns 迭代中的项目数。因此，与上面的示例保持一致，
len(foo) == 4
因此调用
foo[len(foo)]
将加注 IndexError:。这正是您的 encrypt(ptxt) 函数中经常发生的事情
counter = 0
while counter <= len(ptxt):
ctxt = ctxt + exp_ciph(ptxt[counter])
counter += 1
你会想要这样做：
counter = 0
while counter <= len(ptxt) - 1:
ctxt = ctxt + exp_ciph(ptxt[counter])
counter += 1
这是我遇到的前两个问题运行。我不保证没有更多。希望这能让你走得更远。