Predict the output of the following code:
def callon(b=20, a=10): b = b + a a = b - a print(b, "#", a) return b x = 100 y = 200 x = callon(x, y) print(x, "@", y) y = callon(y) print(x, "@", y)

Question

State True or False:
While defining a function in Python, the positional parameters in the function header must always be written after the default parameters.

Answer 1

Initial state: x = 100, y = 200.

Call 1: callon(100, 200)
  b = 100 + 200 = 300
  a = 300 - 200 = 100
  Print: 300 # 100
  Return: 300
  Update x to 300.

Print: x, "@", y → 300 @ 200

Call 2: callon(200) (default a = 10)
  b = 200 + 10 = 210
  a = 210 - 10 = 200
  Print: 210 # 200
  Return: 210
  Update y to 210.

Print: x, "@", y → 300 @ 210

Final Output:
300 # 100
300 @ 200
210 # 200
300 @ 210

Explanation: The function callon accepts two parameters, b and a, with default values of b=20 and a=10. Function logic: - b is updated: b = b + a. - a is updated: a = b - a. - The current values of b and a are printed, separated by "#". - The updated value of b is returned. The initial call to callon modifies the variable x. The second call modifies y. The concluding output reflects the state of x and y after these operations.

Predict the output of the following code:
def callon(b=20, a=10): b = b + a a = b - a print(b, "#", a) return b x = 100 y = 200 x = callon(x, y) print(x, "@", y) y = callon(y) print(x, "@", y)

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Questions Asked in CBSE Class XII exam

Predict the output of the following code: def callon(b=20, a=10): b = b + a a = b - a print(b, "#", a) return b x = 100 y = 200 x = callon(x, y) print(x, "@", y) y = callon(y) print(x, "@", y)

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Solution and Explanation

Top Questions on Commands and Requests

Questions Asked in CBSE Class XII exam

Predict the output of the following code:
def callon(b=20, a=10): b = b + a a = b - a print(b, "#", a) return b x = 100 y = 200 x = callon(x, y) print(x, "@", y) y = callon(y) print(x, "@", y)