为什么这个函数在猜错之后没有重新提示用户？

Question

I have been writing a piece of code in my free time, messing around with opening/closing files to try and get a 100% secure file.

I have this so far:

def StartLock():
    my_pass = open("pass.txt",  "r+")
    passcode = my_pass.read()
    my_pass.close()
    # Establish "passcode" variable by reading hidden text file
    if passcode != "": 
            PasswordLock(input("Password: ")) 
    elif passcode == "": #Passcode is empty
            print("Passcode not set, please set one now.")
            my_pass = open("pass.txt", "r+")
            passcode  = my_pass.write(input("New Pass: ")) #User establishes new pass
            my_pass.close()                
            print("Passcode set to :" + passcode)
            PasswordLock(passcode) #User is passed on with correct pass to the lock

    def PasswordLock(x):
            my_pass = open("pass.txt", "r+")
            passcode = my_pass.read()
            my_pass.close()
            attempts = 3
            def LockMech(x): #Had to do this to set attempts inside instance while not resetting the num every guess
                   if attempts != 3:
                           print("Attempts Left: " + str(attempts))
                   if x == passcode:
                           print("Passcode was correct. Opening secure files...")
                           return True
                   elif attempts == 0:
                           print("You are out of attempts, access restricted.")
                           Close()
                   elif x != passcode and attempts > 0:
                           print("Passcode was not corrent, please try again.") #This does get printed to console when I type in a wrong pass, so it gets here
                           attempts = attempts - 1
                           LockMech(input(":")) #This is what seems to be broken :(

def Close():
    pass

StartLock()

For some reason, when I run this (with a word already stored in "pass.txt") and intentionally type in the wrong password for bug testing, I am not re-prompted to enter another password with my attempts printed as it should.

I have made sure that defining a function inside of another function is acceptable and my spelling is correct, and after playing around with the code trying to get it to work I am not able to find the problem. .

Answer 1

I don't see where LockMech is ever being called.

Having LockMech call itself recursively is an odd way to approach retries. Why not use a while loop or a for loop?

The preferred way to read the passcode is to use a context manager

with open("pass.txt",  "r+") as my_pass:
    passcode = my_pass.read()

This way the file is closed automatically at the end of the block

Here is a simplified version of your PasswordLock . You'll have to make some other small changes to your program to use it

def PasswordLock():
    with open("pass.txt",  "r+") as my_pass:
        passcode = my_pass.read()

    for attempts_remaining in (2, 1, 0):
        x = input("Password: ")
        if x == passcode:
            print("Passcode was correct. Opening secure files...")
            return True
        if attempts_remaining:
            print("Passcode was not corrent, please try again.")
            print("Attempts Left: {}".format(attempts_remaining)

    print("You are out of attempts, access restricted.")
    return False

Notice that I replaced your call to Close() with a return False . Function calls are not the same as "goto". You'll get yourself into bother if you keep trying that