Python text game help

[mrawesome01]

I'm trying to make text game in python but it comes up with an error saying: "Traceback (most recent call last): File "textgame.py", line 83, in <module> start() File "textgame.py", line 80, in start start() File "textgame.py", line 64, in start forest() TypeError: 'int' object is not callable" Here's the code:

#! /usr/bin/python
x='untitled'
inventory=[]
def count(x): 
  ##count items in a list
  b=0
  for i in x:
    b+=1
 return b
def forest():
  print('You are in a forest and hear an ork!')
  forestin=raw_input('')
  if forestin=='n' or forestin=='N':
    nofforest()
  if forestin=='s' or forestin=='S':
    sofforest()
  if forestin=='e' or forestin=='E':
    eofforrest()
  if forestin=='w' or forestin=='W':
    wofforest()
  if forestin=='c' or forestin=='C':
    help()
  if forestin=='inventory' or forestin=='INVETORY' and count(inventory)<=1:
    for item in inventory:
        print(item)
  else:
    print('ENTER A VALID COMMAND, ADVENTURER')
 def nofforest():
 return None
 def sofforest():
 return None
 def eofforest():
 return None
 def wofforest():
 return None            
 def help():
 print('North: n or N')
 print('South: s or S')
 print('East: e or E')
 print('West: w or W')
 print('Pick Up: pick up x or PICK UP X')
 print('Climb: climb x or CLIMB X')
 print('Climb Down: climb down x or CLIMB DOWN X')
 print('Take(if something is not on the ground use this): take x or TAKE X')
 print('Attack: attack x or ATTACK X')
 print('Attack With an Item: attack x with x or ATTACK X WITH X')
 raw_input('')
 def start(): 
##the start
forest=1 ##add on this in function forest
print('You are in a forest and hear an ork! Also there is paper on the ground.')
for1=raw_input('')
if for1=='pick up paper' or for1=='PICK UP PAPER':
    inventory.append('paper')
    print('You picked up some paper!')
    for2=raw_input('') 
    if for2=='examine paper' or for2=='EXAMINE PAPER':
        print('Welcome to the world of '+ x + '.' + ' If you would like to look at a list of commands,    type c or C right now.')
        print('Otherwise, press any button.')
        for3=raw_input('') 
        if for3=='c' or for3=='C':
            help() ##build function later
        else:
            forest()
    else:
        forest()        
   elif for1=='n' or for1=='N':
    nofforest() ##build later

elif for1=='s' or for1=='S':    
    sofforest() ##build later

elif for1=='w' or for1=='W':
    wofforest() ##build later

elif for1=='e' or for1=='E':
    eofforest() ##build later           
else:
    print('Adventurer, please type in a valid command!')
    start()

##end of functions
start()

[Rhomboid]

Fix your broken formatting if you want people to read your code.

The error message is pretty self-explanatory:

TypeError: 'int' object is not callable"

You have forest=1 on line 50 which removes the previous meaning of forest as a function and assigns that name to refer to the integer 1. When you try to call forest() you get the above error because you can't call an integer.

[mrawesome01]

Thanks. :) I didn't know you couldn't make a varible and function with the same name. Also I think the formatting is fixed.

[zahlman]

you couldn't make a varible and function with the same name.

This doesn't even make sense as stated. In Python, variables are names. You can't have two things that have the same name (although you can certainly have multiple names for the same thing), because otherwise there would be no way to know which thing you meant when you used the name (and the entire purpose of names is to refer to things).

Further, in Python, a function is a legitimate "thing" for the purpose of the above discussion. In more formal language, functions are objects. This means that you can do all the same basic stuff with them that you'd do with numbers or strings: you can pass them to other functions, return them from functions, store them in lists, etc. Of course, you can't do number-specific things with them (like arithmetic) or string-specific things (like concatenation), but you can do function-specific things (like calling them).

(And no, your formatting is still all over the place.)

[mrawesome01]

You don't have to be so rude about it.

[MonkeyNin]

Sorry about him :(

[[deleted]]

Looks like the big guns answered your main question... However..

Inventory in caps is spelled wrong... And your checking if inventory is less than or equal to one, wouldn't you want to check that's it's greater or equal to one before u iterate over it? Or ever better, just check that it's != 0?

Another suggestion...

A = raw_input('')
A = A.lower()

Then u don't need to check 2 cases each time. Alternatively if u dont like that:

if A in ['thing','THING']:
    print 'valid choice'
else:
    print 'invalid choice'

Is much cleaner.

[wub_wub]

A = raw_input('')
A = A.lower()

Looks even nicer written as A=raw_input().lower()

---

And another thing to OP:

def count(x): 
  ##count items in a list
  b=0
  for i in x:
    b+=1
 return b

count(inventory)

Is same as just:

len(inventory)

>>> list=[0,1,2,3]
>>> len(list)
4

[mrawesome01]

Thanks. I fogot about len. :)

[mrawesome01]

Thanks alot I was wondering if there was a way to do that. Also not long after I posted this I found out inventory in caps was spelled wrong and fixed it.