How do I write a function with output parameters (call by reference) in Python?

All parameters (arguments) in the Python language are passed by reference. It means if you change what a parameter refers to within a function, the change also reflects back in the calling function.

Achieve this in the following ways −

Return a Tuple of the Results

Example

In this example, we will return a tuple of the outcome −

# Function Definition
def demo(val1, val2):
   val1 = 'new value'
   val2 = val2 + 1
   return val1, val2

x, y = 'old value', 5

# Function call
print(demo(x, y))

Output

('new value', 6)

Passing a mutable object

Example

In this example, we will pass a mutable object −

# Function Definition
def demo2(a):
   # 'a' references a mutable list
   a[0] = 'new-value'
   # This changes a shared object
   a[1] = a[1] + 1

args = ['old-value', 5]
demo2(args)
print(args)

Output

['new-value', 6]

Passing a Dictionary that gets mutated

Example

In this example, we will pass a dictionary −

def demo3(args):
   # args is a mutable dictionary
   args['val1'] = 'new-value'
   args['val2'] = args['val2'] + 1

args = {'val1': 'old-value', 'val2': 5}

# Function call
demo3(args)
print(args)

Output

{'val1': 'new-value', 'val2': 6}

Values in Class Instance

Example

In this example, we will pack up values in class instance −

class Namespace:
   def __init__(self, **args):
      for key, value in args.items():
         setattr(self, key, value)

def func4(args):
   # args is a mutable Namespace
   args.val1 = 'new-value'
   args.val2 = args.val2 + 1

args = Namespace(val1='old-value', val2=5)

# Function Call
func4(args)
print(vars(args))

Output

{'val1': 'new-value', 'val2': 6}

AmitDiwan

Updated on: 19-Sep-2022

4K+ Views

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