# Error on Python

1. Dec 19, 2011

### azerty12

Hi guys
I'm trying to program a Roulette game on Python
My programs asks for inputs and I gave names to these inputs.
The point is, since I didn't assign these names before, the following error occurs:
UnboundLocalError: local variable 'color' referenced before assignment

I tried to get rid of this error with exceptions, but I couldn't
Any idea?

2. Dec 20, 2011

Could you paste your code here? Perhaps it's a problem of local/global variables? Do you refer to a variable which is defined inside a function?

3. Dec 21, 2011

### azerty12

Here's my program (edited with IDLE)
The problem occurs when I execute Roulette() on python's shell and when I choose 'No' when he asks me whether I want to play

I didn't put much comments on my program, so if there is something you don't understand, just ask.

Since I can't attach the file with IDLE, I'm gone send it to you on notepad

#### Attached Files:

• ###### Casino.txt
File size:
1.3 KB
Views:
95
4. Dec 21, 2011

### Staff: Mentor

In your Game function, when the user responds with "No", the col variable is uninitialized.

Code (Text):
def Game(bet,result):
if result==51:
print("first attempt")
else:
print("You have:",bet,"$") if bet==0: return "plucked" play=input("do you want to play? Yes/No") if play=='No': print("Wise decision","you leave with",bet,"$")
elif play=='Yes':
gamble=input("What number do you choose?")
gamble=int(gamble) #Number you bet on
col=color(gamble)
result=randrange(50)

if result==gamble:
return Game(3*bet,result)
elif color(result)==col:
return Game(ceil(0.5*bet),result)
elif gamble>50:
print("wrong choice")
else:
return Game(0,result)
I believe that your problem occurs when the code tries to execute elif color(result) == col.

If this is the problem, you could fix it by assigning a value to col like this:
Code (Text):
def Game(bet,result):
col = 'Black'
if result==51:
print("first attempt")
else:
print("You have:",bet,"$") if bet==0: return "plucked" play=input("do you want to play? Yes/No") if play=='No': print("Wise decision","you leave with",bet,"$")
elif play=='Yes':
gamble=input("What number do you choose?")
gamble=int(gamble) #Number you bet on
col=color(gamble)
result=randrange(50)

if result==gamble:
return Game(3*bet,result)
elif color(result)==col:
return Game(ceil(0.5*bet),result)
elif gamble>50:
print("wrong choice")
else:
return Game(0,result)
Another way that's probably better is to exit the function if the user enters "No".

5. Dec 21, 2011

### azerty12

I'm afraid the first way doesn't work, it generates the following exception:
UnboundLocalError: local variable 'col' referenced before assignment

About your second suggestion: Could you tell me how to exit the function please? (I just begging in programming)

6. Dec 22, 2011

### Staff: Mentor

I didn't test the code I wrote, but I think it should work. Did you notice the line I added to your code?
Code (Text):
def Game(bet,result):

Just return some appropriate value. BTW, there are several code paths in your Game function that don't return anything. Each of the if ... elif branches ought to return something.

7. Dec 22, 2011

### azerty12

Yes I did notice the line you added. But it still raises the same kind of exception.

Do you mean that instead of printing things like "wrong number"... I should return these?

8. Dec 22, 2011

### Staff: Mentor

See if this makes a difference.
Code (Text):
if play=='No':
print("Wise decision","you leave with",bet,"\$")
return
As you have described things, you're getting the error when you type "No". In that case, col doesn't get set, but there is code below that executes, that tries to compare col with color(result).

I think that's what's causing your error.

9. Dec 22, 2011

### jhae2.718

You have quite a few variables that are referenced when they may not be defined. Mark44's suggestion should take care of the problem.

10. Dec 25, 2011

### azerty12

Absolutely Mark, you got it!

Thanks a lot and thx to jihae and radou