CODEWITHSUNDEEP
pythonif-statement
Sam Thomas
Sam Thomas

Reputation: 57

Can someone help me spot logical error Python

I made a simple stack program but I am able to enter more elements than the size given to stack by user input. The if len(self.stk)==size statement doesn't seem to work and I can't understand why. This is my code:

class stack():
    def __init__(self,size):
        self.stk=[]
        self.size = size
    def push(self,item):
        if len(self.stk)==self.size:
            print"OVERFLOW!"
        else:
            self.stk.append(item)
            print "Len of stack is ",len(self.stk)
            print "Size is ",self.size
    def pop(self):
        if self.isempty()==True:
            print "UNDERFLOW"
        else:
            del self.stk[-1]
    def isempty(self):
        if self.stk==[]:
            return True
        else:
            return False
    def display(self):
        print "\nNow going to show you the stack: \n "
        for i in range(len(self.stk)-1,0,-1):
            print self.stk[i]

size = raw_input("Enter size of stack you want: ")
stak = stack(size)
while True:
    choice = int(raw_input("\nEnter \n 1.To push an item \n 2.Pop an item \n 3. Display Stack \n 4.Quit: "))
    if choice == 1:
        elem = raw_input("Enter the element you want to enter: ")
        stak.push(elem)
    if choice == 2:
        stak.pop()
    if choice == 3:
        stak.display()
    if choice==4:
        break

Upvotes: 0

Views: 74

Answers (1)

TemporalWolf
TemporalWolf

Reputation: 7952

You need to typecast the input to a number

size = int(raw_input("Enter size of stack you want: "))

Alternatively, since you're working in Python2.7:

size = input("Enter size of stack you want: ")

Would work, as it will evaluate what they give it and (if given a whole number) will return an integer.

In the future, one thing I may suggest:

Add the following function and you can then inspect all of your variables at any point:

def debug(*args):
    print('Variable debugger output:')
    for arg in args:
        if arg in ["__builtins__", "inspect", "debug"]:
            continue  # skip these variables
        try:
            var = eval(arg)
            print('{0}: {1} {2}'.format(arg, type(var), var))
        # Handles variables not passed as strings
        except (TypeError, NameError):
            print('{0}: {1}'.format(type(arg), arg))

with

debug(*dir())  # All variables in scope

or 

debug('size', 'self.size')  # Specific variables - Note it accepts both a list
debug(size, self.size)      # of strings or a list of variables

which will give you something like:

debug: <type 'function'> <function debug at 0x7fa046a4f938>
in2: <type 'int'> 5
normin: <type 'int'> 23  # normal input
rawin: <type 'str'> 23  # raw input
sys: <type 'module'> <module 'sys' (built-in)>
test: <type 'str'> Hello
testfloat: <type 'float'> 5.234

Note: debug won't show itself if you use the code above... this shows what a function looks like from the debugger.

Upvotes: 3

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