不能杀死我的python代码。发生了什么？

Question

好吧，所以我用python编写了一个非常简单的密码破解程序，它可以强制使用字母数字字符的密码。目前这段代码只支持1个字符的密码和一个包含md5哈希密码的密码文件。它最终将包括一个选项来指定你自己的字符限制（破解程序尝试多少个字符直到失败）。现在我不能在我想让它死的时候杀死它。我已经包括了一个尝试，除了snippit，但它不起作用。我做错什么了？在

代码：http://pastebin.com/MkJGmmDU

import linecache, hashlib

alphaNumeric = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z",1,2,3,4,5,6,7,8,9,0]

class main:
    def checker():
            try:
                    while 1:
                            if hashlib.md5(alphaNumeric[num1]) == passwordHash:
                                    print "Success! Your password is: " + str(alphaNumeric[num1])
                                    break
            except keyboardInterrupt:
                    print "Keyboard Interrupt."

    global num1, passwordHash, fileToCrack, numOfChars
    print "What file do you want to crack?"
    fileToCrack = raw_input("> ")
    print "How many characters do you want to try?"
    numOfChars = raw_input("> ")
    print "Scanning file..."
    passwordHash = linecache.getline(fileToCrack, 1)[0:32]
    num1 = 0

    checker()

main

Answer 1

除了打印，当capturinKeyboardInterrupt时，您需要实际退出程序，您只打印一条消息。在

Answer 2

这对我很有效。。。在

import sys

try:
    ....code that hangs....

except KeyboardInterrupt:
    print "interupt" 
    sys.exit()  

Answer 3

允许KeyboardInterrupt结束程序的方法是不做任何事。它们的工作原理是不依赖于在except块中捕获它们；当异常从程序（或线程）中冒出时，它就会终止。在

您所做的是捕获KeyboardInterrupt并通过打印消息然后继续处理它们。在

至于程序为什么会卡住，没有任何东西会导致num1发生变化，所以md5计算每次都是相同的计算。如果您想迭代alphaNumeric中的符号，那么就这样做：for symbol in alphaNumeric: # do something with 'symbol'。在

当然，这仍然只考虑每个可能的一个字符的密码。你要比那更努力。。。：）

我想你也对类的使用感到困惑。Python并不要求您将所有内容包装在类中。程序末尾的main没有任何用处；代码的运行是因为当编译器试图找出main类是什么时，会对其进行求值。这是滥用语法。您要做的是将这段代码放入main函数，然后调用函数（与当前调用checker的方式相同）。在