Help with defined function and while loop

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January 7th, 2013, 05:42 PM

iam1133

Registered User

Join Date: Jan 2013

Posts: 2

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Help with defined function and while loop

Here is my program so far in Python 2.7:

def menu():
....print "Welcome to my bitchin' calculator!"
....print " "
....return raw_input("Would you like to add, subtract, multiply, or divide? ")
def add(add1,add2):
....print add1, "+", add2, "=", add1 + add2
def sub(sub1,sub2):
....print sub1, "-", sub2, "=", sub1 - sub2
def redo(a):
....if a == "yes":
........print "OK!"
....else:
........loop = 0
loop = 1
choice = 0
while loop == 1:
....choice = menu()
....if choice == "add":
........add(input("add this: "),input("to this: "))
........redo(raw_input("Would you like to start over? "))
print "Bye bye!"

*Note that the .... was added to indicate indention, due to this forum removing my indentions. in the actual program, the indentions are present

I can choose to add, and then it successfully adds the numbers and starts the redo function as defined above. However when I type "no", it should set loop to 0 and stop running the while loop, but it doesn't! after I type no, it runs menu again and starts over. It never prints "Bye bye!" Can someone help a rookie out?

January 7th, 2013, 07:22 PM

b49P23TIvg

Contributing User

Join Date: Aug 2011

Posts: 3,458

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Code:

def redo(a):
    if a == "yes":
        print "OK!"
    else:
        loop = 0    # loop is local to function redo

loop = 1     # this loop has global scope
choice = 0

Please follow the link at my signature to learn about [c o d e] tags. Thank you for the dots! Appreciated.

The easiest solution is

Code:

def redo(a):
    global loop
    if a == "yes":
        print "OK!"
    else:
        loop = 0    # loop is global

A better solution, because redo doesn't need to know details of its environment, is

Code:

def redo(a):
    if a=='yes':
        print('OK!')
        return True
    return False
#...

    loop = redo(raw_input('question').lower())

__________________
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Last edited by b49P23TIvg : January 8th, 2013 at 07:35 AM. Reason: screwy!

January 7th, 2013, 10:10 PM

iam1133

Registered User

Join Date: Jan 2013

Posts: 2

Time spent in forums: 30 m 49 sec
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Thank you, one more question

Thank you for your help! I have one question to your reply. When you type global before any variable, does that mean that the changes made to that variable within that function apply to the program as a whole? and without it, the changes made to that variable won't be applicable to the rest of the program?

January 8th, 2013, 07:45 AM

b49P23TIvg

Contributing User

Join Date: Aug 2011

Posts: 3,458

Time spent in forums: 1 Month 2 Weeks 4 Days 6 h 26 m 43 sec
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global & local scope

The output of these functions depends on the order in which you try them. The interesting experiment is, perhaps,
>>> G
'global'
>>> global_G()
global
>>> G
'changed'

Code:

G = 'global'

def no_assignment_python_finds_G_in_global_scope():
    print(G)

def UnboundLocal_error():
    '''
        G is local because it is assigned to within this function
    '''
    print(G)    # G doesn't yet have a value!
    G = 8       # assigment makes G local


def global_G():
    '''
        G in this function is the global G
    '''
    global G                             # tell python where to find G
    print(G)
    G = 'changed'

Help with defined function and while loop

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