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)

Answer 1

Output: 

300 # 100
300 @ 200
210 # 200
300 @ 210

Explanation: The positional variables x=100 and y=200 supplied are used in the callon() function's initial call. 'a' and 'b' are calculated using these values. Only one of the values of 'y' is supplied into 'b' in the second call to the callon() function, while the default value for 'a' is utilized for the calculation.