Python function accepting an object as a parameter?

Question

So I have the following:

class Tree(object):
    def__init__(self):
        self.right = None
        self.left = None
        self.val = None

I populate the tree with stuff in my code.

I'd like to write a function of the form

def mult(newVal, node)
    newVal = diff(node.left, newVal, var)
    newVal.append('*')
    newval.append(next(node.right))
    newVal.append('+')
    newVal = diff(node.left, newVal, var)
    newVal.append('*')
    newVal.append(next(node.left))

Where next is simply a function that traverses to the next node in the tree, and diff is a recursive function of the following form:

def diff(node, newVal, var):
...
...
    elif(node.val == '*'):
        newVal = diff(node.left, newVal, var)
        newVal.append('*')
        newval.append(next(node.right))
        newVal.append('+')
        newVal = diff(node.left, newVal, var)
        newVal.append('*')
        newVal.append(next(node.left))
...
...

and my "main" is

    node = Tree()
    newEquation = []
    pos = 0
    pos, newTree = buildTree(node, equation, pos)
    newEquation = diff(newTree, newEquation, variable)
    newEquation = ''.join(newEquation)
    print newEquation

How would I write the def mult() function to accept a tree node as a parameter, and then call the diff() function again?

it's fixed, a syntax error was throwing a strange error message. Thanks! — gfppaste
– gfppaste, Commented Apr 14, 2012 at 2:43

Charles Menguy · Accepted Answer · 2012-04-14 02:45:31Z

1

Python is not typed, so if you call mult with:

mult(newEquation, node)

That should work just fine.

answered Apr 14, 2012 at 2:45

Charles Menguy

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