我一直在尝试制作一个加密和解密系统,但我遇到了一个小错误。这是我的代码:
import sys
import pyperclip
def copy(data):
question = input("Copy to clipboard? ")
if question.lower() == 'yes' or question.lower() == 'y':
pyperclip.copy(data)
print("Encrypted message copied to clipboard.")
rerun()
elif question.lower() == 'no' or question.lower() == 'n':
rerun()
else:
print("You did not enter a valid input.")
copy(data)
def rerun():
ask = input("\nWould you like to run this program again? ")
if ask.lower() == "yes" or ask.lower() == "y":
print(" ")
run()
elif ask.lower() == 'no' or ask.lower() == 'n':
sys.exit("\nThank you!")
else:
print("You did not enter a valid input.")
rerun()
def encrypt(key, msg):
encrypted_message = []
for i, c in enumerate(msg):
key_c = ord(key[i % len(key)])
msg_c = ord(c)
encrypted_message.append(chr((msg_c + key_c) % 127))
return ''.join(encrypted_message)
def decrypt(key, encrypted):
msg = []
for i, c in enumerate(encrypted):
key_c = ord(key[i % len(key)])
enc_c = ord(c)
msg.append(chr((enc_c - key_c) % 127))
return ''.join(msg)
def run():
function_type = input("Would you like to encrypt or decrypt a message? ")
if function_type.lower() == "encrypt" or function_type.lower() == "e":
key = input("\nKey: ")
msg = input("Message: ")
data = encrypt(key, msg)
enc_message = "\nYour encrypted message is: " + data
print(enc_message)
copy(data)
elif function_type.lower() == "decrypt" or function_type.lower() == "d":
key = input("\nKey: ")
question = input("Paste encrypted message from clipboard? ")
if question.lower() == 'yes' or question.lower() == 'y':
encrypted = pyperclip.paste()
print("Message: " + encrypted)
elif question.lower() == 'no' or question.lower() == 'n':
encrypted = input("Message: ")
else:
print("You did not enter a valid input.")
run()
decrypted = decrypt(key, encrypted)
decrypted_message = "\nYour decrypted message is: " + decrypted
print(decrypted_message)
copy(decrypted)
else:
print("\nYou did not enter a valid input.\n")
run()
run()
它说 局部变量’encrypted’可能在分配和突出显示之前被引用
decrypted = decrypt(key, encrypted)
在 run() 函数下。
是因为我在其他函数中使用了变量“encrypted”吗?如果是这样,我将如何解决这个问题并仍然保持我的程序的功能?
我对 python 比较陌生,所以如果你能解释你的答案,我将不胜感激。
原文由 rocketsnstuff 发布,翻译遵循 CC BY-SA 4.0 许可协议
是 linter 生成的警告。
这是因为 linter 看到
encrypted
在两个 if 条件下被赋值和
然而,linter 无法知道这两个 if 条件是互补的。因此,考虑到当所有条件都不为真时,变量
encrypted
将最终未初始化。要消除此警告,您可以简单地在任何
if
条件之前初始化变量None
值