import pyperclip, os.path
LETTERS = r""" !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXY
Z[\]^_`abcdefghijklmnopqrstuvwxyz{|}~"""
def main():
message = """Alan Mathison Turing was a British mathematician,
logician, cryptanalyst, and computer scientist. He was highly influential in
the development of computer science, providing a formalisation of the concepts
of "algorithm" and "computation" with the Turing machine."""
mode = 'encrypt'
key = os.urandom(len(message))
if mode == 'encrypt':
translated = encryptMessage(key, message)
elif mode == 'decrypt':
translated = decryptMessage(key, message)
print('%sed message:' % (mode.title()))
print(translated)
pyperclip.copy(translated)
print()
print('The message has been copied to the clipboard.')
def encryptMessage(key, message):
return translateMessage(key, message, 'encrypt')
def decryptMessage(key, message):
return translateMessage(key, message, 'decrypt')
def translateMessage(key, message, mode):
translated = [] # this is gonna store the message string
keyIndex = 0
key = key.upper()
for symbol in message:
num = LETTERS.find(symbol.upper())
if num != -1: # -1 means symbol.upper() was not found in LETTERS
if mode == 'encrypt':
num += LETTERS.find(key[keyIndex]) # Add if encrypting
elif mode == 'decrypt':
num -= LETTERS.find(key[keyIndex]) # Subract if decrypting
num %= len(LETTERS) # Handle the wrap-around
# Add the encrypted/decrypted symbol to the end of translated
if symbol.isupper():
translated.append(LETTERS[num])
elif symbol.islower():
translated.append(LETTERS[num].lower())
keyIndex += 1 # Move to the next letter in the key
if keyIndex == len(key):
keyIndex = 0
else:
# The symbol was not is LETTERS, so add it to translated as it is
translated.append(symbol)
return ''.join(translated)
if __name__ == '__main__':
main()
我正在创建一个通常称为One-Time Pad Cipher的程序,但是当我运行程序时,我得到错误:
TypeError: Can't convert 'int' object to str implicitly*.
程序在第70行停止:
num += LETTERS.find(key[keyIndex]) # Add if encrypting
我该如何解决这个问题?
注意:我知道困扰我的代码的许多其他问题。
答案 0 :(得分:0)
更改if-else条件:
if mode == 'encrypt':
num += LETTERS.find(key[keyIndex]) # Add if encrypting
elif mode == 'decrypt':
num -= LETTERS.find(key[keyIndex]) # Subract if decrypting
到
if mode == 'encrypt':
num += int(LETTERS.find(key[keyIndex])) # Add if encrypting
elif mode == 'decrypt':
num -= int(LETTERS.find(key[keyIndex])) # Subract if decrypting
这是必需的,因为num
类型为int
,而find()
上的LETTERS
正在您的计划中返回str
。
答案 1 :(得分:0)
您的问题是key[keyIndex]
不是字符串。 string.find
方法只接受字符串。
这个程序在python 2.7中运行得很好,但在python 3.2中,这就是造成这个问题的原因。
在key[keyIndex]
中包裹str()
。
if mode == 'encrypt':
num += LETTERS.find(str(key[keyIndex]))
elif mode == 'decrypt':
num -= LETTERS.find(str(key[keyIndex]))
你甚至可以创造一种让生活更轻松的功能。
def findKeyValue(alphabet, key, index):
return alphabet.find(key[index].__str__())
则...
if mode == 'encrypt':
num += findKeyValue(LETTERS, key, keyIndex)
elif mode == 'decrypt':
num -= findKeyValue(LETTERS, key, keyIndex)