Programming fundamentals & OOP (classes, inheritance, polymorphism) – A-level Computer Science exam questions with mark scheme

1Core3 marks

(a) State what is meant by encapsulation in object-oriented programming. [1]

(b) Explain two differences between a local variable and a global variable. [2]

(a) Combining/bundling the data (attributes) and the methods that act on that data into a single object/class; accept: hiding an object's internal state so it can only be accessed through its methods(1)
(b) A local variable is only accessible/has scope within the subroutine/block in which it is declared, whereas a global variable is accessible throughout the whole program(1)
(b) A local variable only exists (is allocated memory) while its subroutine is executing, whereas a global variable exists for the whole time the program runs; accept: a local variable with the same name as a global variable takes precedence inside that subroutine(1)
Answer: (a) Bundling data and the methods that operate on it into one class/object (with internal state hidden). (b) Scope: local = only within its subroutine, global = whole program; Lifetime: local exists only while subroutine runs, global for the whole run.
Examiner tip: For 'explain two differences' each mark needs both sides of the comparison – saying only what a local variable does is not a difference.

2Standard4 marks

The algorithm below is written in pseudo-code.

n ← 2026
total ← 0
WHILE n > 0
   total ← total + (n MOD 10)
   n ← n DIV 10
ENDWHILE
OUTPUT total

(a) Complete a trace table, with columns n and total, for the algorithm. Show the initial values in the first row. [3]

(b) State the purpose of the algorithm. [1]

(a) n column correct: 2026, 202, 20, 2, 0(1)
(a) total column correct up to and including 8: 0, 6, 8, 8(1)
(a) Final value of total is 10 (and no extra rows); ignore an OUTPUT column if 10 is shown(1)
(b) It calculates/outputs the sum of the digits of n (the starting number); do not accept 'adds up n'(1)
Answer: (a) n: 2026, 202, 20, 2, 0; total: 0, 6, 8, 8, 10. (b) Outputs the sum of the digits of the number.
Examiner tip: MOD 10 gives the last digit and DIV 10 removes it – students who use ordinary division get 202.6 and lose every later row.

3Standard5 marks

The recursive function F is shown below.

FUNCTION F(a, b)
   IF b = 0 THEN
      RETURN a
   ELSE
      RETURN F(b, a MOD b)
   ENDIF
ENDFUNCTION

(a) State the value returned by the call F(84, 36). [1]

(b) State how many times F is called in total, including the original call, when F(84, 36) is executed. [1]

(d) Identify the base case in F and explain why a recursive subroutine must have a base case. [2]

(a) 12(1)
(b) 3 (calls F(84, 36), F(36, 12), F(12, 0))(1)
(c) The highest common factor / greatest common divisor of a and b; accept HCF / GCD; NE 'a factor of a and b'(1)
(d) Base case: when b = 0 (the function returns a without calling itself); accept IF b = 0 THEN RETURN a(1)
(d) Without a base case the subroutine would keep calling itself and never return / terminate; each call adds a stack frame, so eventually the stack overflows / runs out of memory (and the program crashes); accept: the base case is the stopping condition that allows the calls to unwind and return values(1)
Answer: (a) 12 (b) 3 (c) The HCF/GCD of a and b (d) Base case is b = 0 (return a); without one the function keeps calling itself, never returns, and stack frames build up until the stack overflows.
Examiner tip: In (d) say what would actually happen without a base case (endless calls, stack frames building up until stack overflow) rather than just 'it loops forever'.

4Standard6 marks

The Python program below defines two classes.

class Vehicle:
    def __init__(self, reg, wheels):
        self.__reg = reg
        self._wheels = wheels

    def describe(self):
        return self.__reg + " has " + str(self._wheels) + " wheels"

class Lorry(Vehicle):
    def __init__(self, reg, wheels, maxLoad):
        super().__init__(reg, wheels)
        self.__maxLoad = maxLoad

    def describe(self):
        return super().describe() + ", max load " + str(self.__maxLoad) + " tonnes"

l = Lorry("KX70 ABC", 6, 26)
print(l.describe())

(a) Identify the name of the subclass. [1]

(b) Identify a method that has been overridden. [1]

(d) The attribute __reg is private. Explain why a programmer would make an attribute private and how other objects could still be allowed to read its value. [2]

(a) Lorry(1)
(b) describe; accept __init__ / the constructor (also redefined in Lorry)(1)
(c) KX70 ABC has 6 wheels(1)
(c) , max load 26 tonnes – i.e. full output KX70 ABC has 6 wheels, max load 26 tonnes on one line; ignore missing quotation marks(1)
(d) So that it cannot be accessed/changed directly by code outside the class, preventing accidental/invalid changes (encapsulation/information hiding); NE 'for security' on its own(1)
(d) A public getter/accessor method (e.g. getReg) could be provided that returns the value; accept: a public property/read-only property(1)
Answer: (a) Lorry (b) describe (c) KX70 ABC has 6 wheels, max load 26 tonnes (d) Stops other code changing it directly/invalidly; provide a public getter method.
Examiner tip: In (c) follow super().describe() into the parent class first – many students output only the Lorry part.

5Standard6 marks

Write a program that:

repeatedly asks the user to enter a password until the password entered is valid
a valid password has at least 8 characters and contains at least one digit (0–9)
displays a suitable message each time an invalid password is entered
once a valid password has been entered, displays the number of attempts the user took.

You may write your answer in pseudo-code or a programming language you have studied. [6]

Gets a password from the user inside an indefinite loop (WHILE / REPEAT…UNTIL / while True with break)(1)
Checks the length of the password is ≥ 8 (accept > 7)(1)
Checks whether at least one character is a digit, by iterating through the characters or with a suitable built-in (e.g. isdigit, any(), character code between 48 and 57)(1)
Loop ends only when both conditions are true (correct use of AND / correct loop condition); R. if a password meeting only one condition is accepted(1)
Displays an error message for each invalid password, and not for the valid one(1)
Attempts counter initialised before the loop, incremented once per attempt and output after the loop ends (gives 1 if the first password is valid)(1)

Answer: e.g. (Python)

attempts = 0
valid = False
while not valid:
    pw = input("Password: ")
    attempts += 1
    hasDigit = False
    for c in pw:
        if c.isdigit():
            hasDigit = True
    if len(pw) >= 8 and hasDigit:
        valid = True
    else:
        print("Invalid password")
print("Attempts:", attempts)

Examiner tip: Check the counter: initialising it inside the loop or incrementing it only on invalid attempts gives a wrong count and loses the final mark.

6Hard4 marks

Consider the Python program below.

def change(x, items):
    x = x + 1
    items.append(x)

a = 5
b = [1]
change(a, b)
print(a, b)

(a) State the output of the program. [2]

(b) Explain why the value of a and the contents of b are affected differently by the call to change. [2]

(a) 5 (a is unchanged)(1)
(a) [1, 6] (b now contains 6)(1)
(b) x is a local variable that receives the value of a; x = x + 1 only changes the local x (integers are immutable, so x now refers to a new value 6), and a in the main program stays 5; accept: a is (effectively) passed by value / a copy of its value is passed, so changing x does not change a(1)
(b) items refers to the same list object as b (a reference to the list is passed, so changes made through items are seen through b); append changes that list in place, so b becomes [1, 6]; accept: b is (effectively) passed by reference so the change made inside the subroutine affects b(1)
Answer: (a) 5 [1, 6] (b) x is effectively a local copy of a's value; items references the same mutable list as b, so append changes b.
Examiner tip: In Python both arguments pass a reference to an object: say what happens to each – x = x + 1 makes the local x refer to a new integer (a is unchanged), while append changes the shared list in place (so b changes).

7Stretch6 marks

The class diagram shows part of a drawing application. Shape is an abstract class; its method area is abstract (shown in italics).

(a) State the type of relationship between Drawing and Shape shown by the hollow diamond, and explain what it means for the lifetime of the Shape objects. [2]

(b) Explain why area has been declared as an abstract method in Shape rather than being implemented there. [2]

(c) The attribute shapes is a list of Shape objects. Write the method totalArea for the Drawing class so that it returns the total area of all the shapes in the drawing. Your method must make use of polymorphism. You may use pseudo-code or a programming language you have studied. [2]

(a) Aggregation (association aggregation)(1)
(a) The Shape objects can exist independently of the Drawing / are not destroyed when the Drawing object is destroyed (unlike composition)(1)
(b) There is no general way to calculate the area of an unspecified shape, so Shape cannot provide a sensible implementation(1)
(b) It forces every (concrete) subclass to provide/override its own area method, so any Shape object can be asked for its area; accept: so that area() can be called polymorphically on any object in the list of shapes(1)
(c) Iterates over every object in shapes, accumulating a total (initialised to 0) which is returned(1)
(c) Calls area() on each object without checking its class (e.g. no IF type = Circle), so the correct overridden method is used for each object(1)
Answer: (a) Aggregation – shapes can exist after the drawing is deleted. (b) The area can't be calculated for a general shape; every subclass must override area. (c) e.g.
```
def totalArea(self):
    total = 0
    for s in self.shapes:
        total += s.area()
    return total
```
Examiner tip: Using IF statements to test which subclass an object is defeats polymorphism and loses the final mark in (c).

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