Question

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)
    

Hint:

Understand the sequence of operations and how variables are updated during function calls to accurately predict the output of a Python program.

Answer 1

# Step-by-step execution:
# 1. Initially, x = 100 and y = 200.
# 2. First function call: callon(100, 200)
#    b = 100 + 200 = 300
#    a = 300 - 200 = 100
#    Output: 300 # 100
#    Return value: 300
#    x is updated to 300.

# 3. print(x, "@", y)
#    Output: 300 @ 200

# 4. Second function call: callon(200)
#    b = 200 + 10 = 210
#    a = 210 - 10 = 200
#    Output: 210 # 200
#    Return value: 210
#    y is updated to 210.

# 5. print(x, "@", y)
#    Output: 300 @ 210

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

Explanation: The function callon takes two arguments b and a, with default values b=20 and a=10. Inside the function: - b is updated as b = b + a. - a is updated as a = b - a. - The function prints the values of b and a, separated by "#". - Finally, the updated value of b is returned. The first function call updates x, while the second function call updates y. The final output is generated based on the updated values of x and y.