[Type B] Chapter– 3 Class 12 CS – Sumita Arora Assignment | Q/A

Solution:

Indentation and invalid keyword selection that means syntax error is present in the both cases.

(a) Corrected Code
total =0
def sum(arg1, arg2):
    total = arg1+arg2
    print("Total :",total)
    return total
sum(10,20)
print("Total :",total)

Output:
Total : 30
Total : 0

(b) Corrected Code
def Tot(Number):
    Sum = 0
    for C in range(1, Number+1):
        Sum+=C
    return Sum
print(Tot(3))
print(Tot(6))

Output:
6
21

Solution:

Firstly the line 9 will execute and a variable with value 5 will be initialized.
After that in line 10 calcSquare(n) will be called and the respective function will execute means Line 5 will be called.
Then on moving line 6 power function will call and the flow jump to line 1 and result of power function stored in variable a.
Then flow will jump to the line 6 and then 7 and then to line 10 to initialize the result variable.
Then, the Line 11 will be executed to provide output.

Flow cycle is 
9 → 10 → 5 → 6 → 1 → 2 → 3 → 7 → 10 → 11

def addEm(x, y, z):
    print(x+y+z)

Solution : This function will return nothing as of no return keyword.

def addEm(x, y, z):
    return (x+y+z)
    print(x+y+z)

Solution:
This function will print depend on case:
If x, y, z will be numeric value than a numeric value will be printed.
If string will be passed as argument than a string concatenation will be result.

Solution:

(a) 
1
1
1

(b)
1
10
1

(c)
1
10
10

(d)
Hello there!

a =10
y = 5
def myfunc():
    y=a
    a=2
    print("y=", y, "a=", a)
    print("a+y=", a+y)
    return a+y
print("y=", y, "a=", a)
print(myfunc())
print("y=", y, "a=", a)

Output:
y= 5 a= 10
UnboundLocalError: local variable 'a' referenced before assignment

Note: The unbound error occur because during call of myfunc() local variable is used before initializing the same in local scope.

def addEm(x,y,z):
    return x+y+z
    print("the answer is", x+y+z)

The print command is initialized after return command and return command is termination of any function call. Hence, the print() command will never run in such call.

def fun():
    return None

def multiply(number1, number2):
    answer = number1*number2
    print(number1, 'times', number2, "=", answer)

    return(answer)

output = multiply(5,5)

(i) When the code above is executed, what prints out?
(ii) What is variable output equal to after the code is executed?

(i) Output:
5 times 5 = 25
(ii)  output = 25

def multiply(number1, number2):
    answer = number1*number2
    return(answer)
    print(number1, 'times', number2, "=", answer)

output = multiply(5,5)

(i) When the code above is executed, what prints out?
(ii) What is variable output equal to after the code is executed?

(i) Output:
Nothing
(ii) output = 25

Solution:

(a)Correct code:
def minus(total, decrement):    # semicolon was not given
    output = total-decrement
    print(output)
    return output

(b) Correct code:
def check():      # define → def  and semicolon absent
    N =input('Enter N:')
    i =3
    answer =1+i**4/N
    return answer    # Return  → return

(c) Correct code

def alpha(n, string='xyz', k=10):
    return beta(string)
    return n

def beta(string): #  Semicolon absent
    return string == str(n)

print(alpha("Valentine Day"))  # wrong semicolon present
print(beta(string='true'))
print(alpha(n=5, "Good bye"))  # wrong semicolon present

1. def sum(a, b, c, d):
2.     result = 0
3.     result =result+a+b+c+d
4.     return result
5.
6. def length():
7.     return 4
8.
9. def mean(a, b, c, d):
10.    return float(sum(a, b, c, d))/length()
11.
12. print(sum(a, b, c, d), length(), mean(a, b, c, d))

Solution:

Till line 10 is called means a local environment being created and all the three function sum(), length(), mean() being called as by line 12 respectively. 

Which shows that from the sum() variable a, b, c, d will globally access by function and acted upon local variable result and returned.

In case of length() no variable acceptance and declaration will be done. 

In case of mean() variable a, b, c, d will globally access by function and calculated portion will return.

12 → 1 → 2 → 3 → 4 → 12 → 6 → 7 → 12 → 9 → 10

def func1():
    a=1
    b=2

def func2():
    c=3
    d=4

e=5

Variable a and b having same local scope for function func1().
Variable c and d having the same local scope for function func2().

def follow_num(n):
    return n+1

follow_num(4)
follow_num(6)
follow_num(8)
follow_num(2+1)
follow_num(4-3*2)
follow_num(-3-2)

Output:
5
7
9
4
-1
-4

# code
1. def follow_num(n):
2.     return n+1

#non-void code
3.  def follow_num_non(n): 
4.      print(n+1)
5.      return n+1
6.
7.  follow_num(5)
8.  follow_num_non(4)

Flow control
7 → 1 → 2 → 8 → 3 → 4 → 5 

(i)
def increment(n):
    n.append([4])
    return n
L = [1,2,3]
M =increment(L)
print(L,M)


(ii)
def increment(n):
    n.append([49])
    return n[0], n[1], n[2], n[3]
L = [23,35,47]
m1, m2, m3, m4 =increment(L)
print(L)
print(m1, m2, m3, m4)
print(L[3] == m4)

output:
(i)  
[1, 2, 3, [4]] [1, 2, 3, [4]]

(ii) 
[23, 35, 47, [49]]
23 35 47 [49]
True