The second rule for If-Statements here, which has me confused states that:
If this else should never run because it doesn't make sense, then you must use a die function in the else that prints out an error message and dies, just like we did in the last exercise. This will find many errors.
Here's the code from the last exercise:
def dead(why):
print why, "Good job!"
exit(0)
def start():
print "You are in a dark room."
print "There is a door to your right and left."
print "Which one do you take?"
choice = raw_input("> ")
if choice == ‘left’:
bear_room()
else:
dead(‘You stumble around the room until you starve.’)
Is it essentially saying that you must successfully terminate the program if a condition is not met?
Yes, the idea is:
import sys
def die(msg):
print msg
sys.exit(1)
if condition:
# do stuff
else:
die('This cannot happen!')
You could also use an assert instead, or raise an exception, or anything else that would fail catastrophically. This helps you validate at runtime that the clause you didn't expect to execute, really didn't run.
IMHO you shouldn't get too hung up on how this die is done, exactly. The important point the referred text tries to make is that when you're sure some condition is true, you might as well assert it forcefully so that you can catch runtime bugs.
The guide is trying to show you that you can see the choice that a user gives when it's something that you don't check for in your if statement. If your else looks like:
else:
dead(choice)
You will be able to see what the user input that you didn't expect.
The problem is that the syntax only allows for "Left", when the prompt allows entry of anything. In other words, if I enter "Bob", I will starve. I should be trapping for anything that isn't appropriate and "die"ing at that point (or helping the user make an appropriate decision).
This is an attempt at teaching error handling and die. You would rewrite it such that you allow accurate entry or die (no exit). If you don't choose left, there is no other choice even though the prompt states there are two choices.
If you truly expect only 'left' to be added, a better way of checking for unexpeted input would be
if choice != ‘left’:
dead(‘You stumble around the room until you starve.’)
bear_room()
That way you still validate that the input is what you expected, but it saves indentation and space for the main logic.
Related
I am making a small text-based game in Python. It involves many inputs and so to avoid bugs, there are a few things I have to check every time an input exists. Naturally, to speed up the process I wanted to put the code into a def in order to simplify the writing process. When I put the code in the def, it red underlines the continue and break commands (meaning they are incorrect), and if you run the code using the def name, a Traceback occurs. I have tried putting the def section at the beginning of the program, after the while True: (The program is supposed to run infinitely until a certain action is taken that breaks the loop) I have also made sure to try putting it under any variables referenced and in the loop so that no part of it is not defined and so that everything would work if I were to just put the code in there.
Here is the code I am trying to put into a def.
def input_bugs():
if letter_one.lower() == "done" and total_count == 0:
print("You have to play at least one game!")
continue
elif letter_one.lower() == "done":
break
elif len(letter_one) > 1:
print("Sorry, you gotta pick a single letter, no more. Otherwise, type 'done' to end the game and see your stats.")
continue
Here is the Traceback I get every time I try to run it.
line 20
continue
^^^^^^^^
SyntaxError: 'continue' not properly in loop
At this point, I don't even care if I have to write it out every time, I can just copy and paste. I am simply curious as to why it doesn't work so that I know in the future. In case you could not tell, I am pretty new to programming so I want to learn from any mistake I make. Also sorry if I referred to some things in the wrong way. Hopefully, you understood what I meant.
I am VERY new to coding in general so this may come across as a stupid question.
Right now I am experimenting with conditional statements and trying to code a little decision based two-way path in the style of a text game, where the user is given two options and each option has a different outcome. Right now, the "if True" string is ALWAYS executed in the console, regardless of what the user types in. Why is this happening and how can I rewrite it to make the command happen correctly? Also, I thought I followed the instructions for pasting code here but it doesn't look quite right.
input ("Type 'left' if you want to go left and 'right' if you want to go right"
right = True
left = False
right = "right"
left = "left"
if True:
print("You have been eaten by a monster. Game Over!")
else:
print("Congratulations you have made it to the castle!")
else:
print("Error")```
There's no such thing as a stupid question!
In any programming language, when you come across an if-statement, you should read it as "if this statement evaluates to true, do this." So, in your example, writing
if True:
# do something
will always run that code, because True always evaluates to True!
It sounds like what you want to do is compare the player's input to some set string (in this case, "right" or "left"). The first thing you should know is that the input() function returns, as a string, whatever the user input before hitting enter. So, in this case, you might want to do something like
direction = input("Would you like to go left or right?")
Now, once the user inputs something and hits enter, whatever they entered will be saved in the variable direction. Now, we can compare the entered string to whatever we want. In this example, we would want something like
if direction == "left":
# do left stuff
elif direction == "right":
# do right stuff
else:
# they didn't enter a valid string!
I hope this helps. If you're just starting out on your programming journey, I would recommend looking up some Python tutorials online which are specifically made for beginners, and try finding one that suits your learning style. Best of luck!
I would like to know why this code does not work; it should exit at the "GAME OVER" point, but it continues to my next defined function.
I have tried other variations on exit() such as: sys.exit(), quit() and SystemExit.
run_attack = input("What do you do: Run/Attack\n")
run = ['run', 'Run', 'RUN']
attack = ['attack', 'Attack', 'ATTACK']
run_attack = 1
while run_attack < 10:
if run_attack == ("run") or ("Run") or ("RUN"):
print ("You turn to run from the wolf but he quickly pounces
you...")
time.sleep(2)
print("You are quickly ripped apart and just about get to see
yourself be eaten.")
print("GAME OVER")
break
exit() #This is where the game should exit, yet after input it
continues to the next function
elif run_attack == ("attack") or ("Attack") or ("ATTACK"):
print("You brace yourself for a bite and have no time to reach"
"for any kind of weapon form your backpack.")
time.sleep("2")
input("You clock the dog hard, twice on the muzzle.")
print("The dog recoils in pain and retreats back to the woods.")
print("You quickly start running as you assume there will be a den in the woods.")
break
else:
input("Type Run or Attack...")
You have several problems in your code; why did you write this much without testing it?
First, you read the user's input, immediately replace is with 1, and then try to test it (incorrectly) as if it were still a string. Your posted code has several syntax errors, so I have some trouble reproducing the problem. However, the immediately obvious problem is here:
break
exit() # This is where ...
You can't get to the exit statement, as you break from the loop just before you can get there.
I strongly recommend that you back up to a few lines and use incremental programming: write a few lines at a time, debug those, and don't continue until they do what you want.
Also look up how to test a variable against various values. Your if statement is incorrect. Instead, try the list inclusion you're trying to set up:
if run_attack in run:
...
elif run_attack in attack:
...
I took the liberty of rewriting your whole program to show you a few things wrong with it and a few tricks. I've done it without the loop, since you never use it anyway... you can add the while loop later once you've mastered it, but you should really go back to basics on some things here:
run_attack = input("What do you do: Run/Attack\n")
if run_attack.lower() == "run":
print("""some
stuff
with
multiple
lines and GAME OVER""")
exit()
elif run_attack in ("attack", "Attack", "ATTACK"):
print("""some
stuff
with
multiple
lines""")
else:
input("Type Run or Attack...")
Some notes:
Using """ for strings enables you to write multiple lines without multiple print statements
Using str.lower() on strings makes everything easy to compare because you only have to compare it to the lowercase version of each string. However for attack you can notice I used a different inclusion test, without multiple conditions. Either way works here.
Like the other answer here (and many comments), you should use only exit() to leave the program entirely, or only break to exit the loop and continue to other code that's beneath the entire loop.
When you rewrite your loop, with a condition like while number_of_turns < 10 don't forget to add 1 to the number of turns on each loop, otherwise that condition is always True and you'll have an infinite loop...
I'm actually quite surprised this code had any resemblance to the behavior you expected from it, my suggestion is to go back over to the basics of python, learn loops, string methods, basic commands. The rest is already said in the other answer here (which is better than mine, frankly) just wanted to add some ideas.
1.I'm using break to break out of a loop, but I don't know how to make the program keep going no matter what unless this happens. Just typing in while: is invalid (or so the progam tells me) and I want the game to keep going even if the user types in an emptry string.
2.Is there a way to not have to re-type a bit of code every time I need it? I have a bunch of responses for the program to spit out that I'll have to use many times:
if action[0]=='go':
print("You're supposed to go to David!")
elif action[0]=='look':
print("You can't see that")
elif action[0]=='take':
print("You don't see the point in taking that.")
else:
print("I don't recognise that command")
Where action is a list from the player's input. Or do I just have to type it out again each time?
I don't know how to define a function that does the above, and I'm not even sure that's what I'm supposed to do.
3.Some story descriptions I'm using are a very long stings and I don’t want players to have to scroll sideways too much. But I want to just define them as variables to save myself some typing. Is there a way around this. Or do I just have to type it out every time with
print(“““a string here”””)
4.If the string starts with 'look' and has 'floor' or 'mess' or 'rubbish' in it, I want it to print a certain output. This is what I currently have:
if action[0]=='look':
if 'floor' in action or 'rubbish' in action or 'trash' or 'mess' in action:
print('onec')
elif 'screen' in action or 'computer' in action or 'monitor' in action:
print('oned')
elif 'around' in action or 'room' in action or 'apartment' in action:
print('onee')
elif 'david' in action or 'tyler' in action or 'boy' in action or 'brat' in action or 'youth' in action:
print('onef')
break
else:
print("You can't see that")
It prints 'onec' for any input beginning with 'look'.
The while statement requires a condition.
You can call the same instructions over and over using a function.
"String literals can span multiple lines in several ways"
Use strategically-placed print statements to show the value of action, e.g. after if action[0]=='look'
Lastly, please don't add any more items to this question. Rather ask a new question. This site has somewhat specific rules on that sort of thing.
To make an infinite While loop, use while True:.
You could use a dict to store common action strings and their responses.
Just register the string first, then when the input come, change it:
command = "nothing"
command = input("Enter command: ")
while command:
Or just simply:
while True:
Yes, think about it by yourself.. Okay, why not put it in list responses?
If it is really long, put it in a file. Read it when you need it using open(). More on File Processing
This will help you shorten your code, making it easier to read, and makes it more efficient.
while requires a condition that it has to evaluate. If you want it to loop forever, just give it a condition that always evaluates to True, such as 4>3. It would be best for everyone if you just used while True:, which is the clearest option.
For this specific case, I would recommend using a dict() and its .get() method. Something like this:
action_dict = {'go':"You're supposed to go to David!",
'look':"You can't see that",
'take':"You don't see the point in taking that."
}
print(action_dict.get(action[0], "I don't recognise that command")
would replicate what you have going on right now.
See the link provided by cjrh here: http://docs.python.org/3.3/tutorial/introduction.html#strings
Our mind-reading powers are a bit off in October, we'll need some more information other than "it does not work" to help you with that.
Okay, so I'm writing a very simplistic password cracker in python that brute forces a password with alphanumeric characters. Currently this code only supports 1 character passwords and a password file with a md5 hashed password inside. It will eventually include the option to specify your own character limits (how many characters the cracker tries until it fails). Right now I cannot kill this code when I want it to die. I have included a try and except snippit, however it's not working. What did I do wrong?
Code: 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
The way to allow a KeyboardInterrupt to end your program is to do nothing. They work by depending on nothing catching them in an except block; when an exception bubbles all the way out of a program (or thread), it terminates.
What you have done is to trap the KeyboardInterrupts and handle them by printing a message and then continuing.
As for why the program gets stuck, there is nothing that ever causes num1 to change, so the md5 calculation is the same calculation every time. If you wanted to iterate over the symbols in alphaNumeric, then do that: for symbol in alphaNumeric: # do something with 'symbol'.
Of course, that will still only consider every possible one-character password. You're going to have to try harder than that... :)
I think you're also confused about the use of classes. Python does not require you to wrap everything inside a class. The main at the end of your program does nothing useful; your code runs because it is evaluated when the compiler tries to figure out what a main class is. This is an abuse of syntax. What you want to do is put this code in a main function, and call the function (the same way you call checker currently).
Besides printing, you need to actually exit your program when capturin KeyboardInterrupt, you're only printing a message.
This is what worked for me...
import sys
try:
....code that hangs....
except KeyboardInterrupt:
print "interupt"
sys.exit()
Well, when you use that try and except block, the error is raised when that error occurs. In your case, KeyboardInterrupt is your error here. But when KeyboardInterrupt is activated, nothing happens. This due to having nothing in the except part. You could do this after importing sys:
try:
#Your code#
except KeyboardInterrupt:
print 'Put Text Here'
sys.exit()
sys.exit() is an easy way to safely exit the program. This can be used for making programs with passwords to end the program if the password is wrong or something like that. That should fix the except part. Now to the try part:
If you have break as the end of the try part, nothing is going to happen. Why? Because break only works on loops, most people tend to do it for while loops. Let's make some examples. Here's one:
while 1:
print 'djfgerj'
break
The break statement will stop and end the loop immediately unlike its brother continue, which continues the loop. That's just extra information. Now if you have break in a something like this:
if liners == 0:
break
That's going to depend where that if statement is. If it is in a loop, it is going to stop the loop. If not, nothing is going to happen. I am assuming you made an attempt to exit the function which didn't work. It looks like the program should end, so use sys.exit() like I showed you above. Also, you should group that last piece of code (in the class) into a seperate function. I hope this helps you!