Changing a variable in a function

[Danthier]

Pretty new to Python, but I have some background in C and VB. Just joined this site because I couldn't find anywhere else to get Python help!

I'm working with VPython in the IDLE environment, and I'm having a problem changing a variable using a toggle switch and a function. Here's the necessary parts:

mode = 0
def switchMode():
    if t1.value:
        mode = 1
    else:
        mode = 0

tMoGen = toggle(pos=(40,-30),width=10,height=10,text0='Off',text1='On',action=lambda: switchMode())

while True:
    rate(5)
    c.interact()
    print mode

Mode just stays at 0, no matter how many times I flip the switch.
I think I may need to return the value out of the function, but I'm not sure how to do that yet.

[jcao219]

You're right, you need return.

Python has local and global scopes, functions have their own local scope, so a local variable is changed inside of the function, it doesn't affect a variable outside the function's scope.

Add return mode to switchMode(),
and then whenever you call switchMode,
do this: mode = switchMode()

[Danthier]

That was an easy fix, thanks!

[ultimatebuster]

The mode declared is outside the function, which makes it a GLOBAL variable.

The mode within the function is a LOCAL variable, which doesn't affect the mode that's declared outside of the function

A simple fix would be to declare that the mode inside the function is a global variable.

mode = 0
def switchMode():
    global mode # This tells python that all reference to "mode" is global within the current function.
    if t1.value:
        mode = 1
    else:
        mode = 0

The wording is probably really weird, so I'm going to try to explain via example:

var1 = 88
print var1 # 88 would be print in console
function() # Executes function(), which will error out, of course


def function():
    print var1 # This would print 88, as python didn't find a local variable
    var1 += 1 # This is suppose to add 1 to var1, however you will get "UnboundLocalError: local variable 'var1' referenced before assignment" 
    # This is because there's no local variable named "var1". Since python evaluates the right side of an equation first, and that you are creating a new local variable after the evaluation, python gets confused and errors out.

var1 = 88 # Declare this again
print var1 # will print out 88 again
function() #Executes the function
print var1 # will print out 88, not 90. You can see from this that var1 outside is completely different from the var1 that's inside function().

def function():
    var1 = 90 # You might think that you changed var1 that's declared outside of this function to 90, but really what you did is that you created a new variable within the function and set it to be 90
    print var1 # will print 90 as expected
    print 90

To fix these problems, we need the global statement

var1 = 88
print var1 # prints out 88
function() # Will not error out
print var1 # Will print out 91 this time around, instead of 88

def function():
    global var1 # Tells python that var1 referenced insdie the function is a global variable, not a local one
    var1 += 1 # adds 1 to the GLOBAL var1, which is 89
    print var1 # prints out 89
    var1 = 91 # sets the GLOBAL var1 to be 91
    print var1 # prints out 91

[jcao219]

Never use the global function, though.
Use of it is a sign of a bad program structure.

[ultimatebuster]

Not sure if I agree with that. I think they are great temporary variables there's needed for multiple classes until they are passed off as an argument to something else.

[jcao219]

It has to be made clear that global should never be used.

Whenever you see the need to write a global statement, I strongly encourage you to rewrite some parts of your code.

[Danthier]

I do have enough background in C to know what a global variable is, I just didn't even think about that for here, nor did I know the syntax.

I've also done enough Actionscripting to know that I tend to get SERIOUSLY CONFUSED when I use too many globals, so I think I'll get into the habit of just returning a value, now that I know how to do that.