Question

#it's python 3.2.3
class point:
    def __init__(self, x, y):
        self.x = x
        self.y = y

    def __eq__(self, point):
        return self.x == point.x and self.y == point.y

    def __str__(self):
        return 'point(%s, %s)' % (self.x, self.y)

def someFunc(point):
    if point.x > 14: point.x = 14
    elif point.x < 0: point.x = 0

    if point.y > 14: point.y = 14
    elif point.y < 0: point.y = 0

    return point

somePoint = point(-1, -1)
print('ONE: %s' % somePoint)
if somePoint == someFunc(somePoint):
    print('TWO: %s' % somePoint)

I see that there is no somePoint variable assignment after first print()
but variable somePoint magically changes after if statement

Output of this program should be

ONE: point(-1, -1)

But it is

ONE: point(-1, -1)
TWO: point(0, 0)

Could anybody explain me why somePoint changes after

if somePoint == someFunc(somePoint):

condition?

p.s. sorry if my english is bad

Answer 1

You change the value of point inside function someFunc when you call it
in your if-statement, so I would expect the value to be (0,0) at the
end. The reason is that you are passing a reference (or “Pass By Object Sharing”) to the function and any changes to it are reflected later on. This is unlike the pass-by-value method were a local copy is automatically produced.

To avoid changing the original point being passed in you could create a local variable inside someFunc.

Something like this:

def someFunc(a_point): # note new parameter name

    loc_point = point(a_point.x, a_point.y)  # new local point

    if loc_point.x > 14: loc_point.x = 14
    elif loc_point.x < 0: loc_point.x = 0

    if loc_point.y > 14: loc_point.y = 14
    elif loc_point.y < 0: loc_point.y = 0

    return loc_point

Also, it’s probably best not to use point to both refer to your class and your parameter.