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Programming and Software Development: Question 3

Syllabus 11.3

Structured AS 7 marks

A programmer writes this procedure, intending it to swap the values held in two variables in the main program.

PROCEDURE Swap(BYVAL A : INTEGER, BYVAL B : INTEGER)
    DECLARE Temp : INTEGER
    Temp ← A
    A ← B
    B ← Temp
    OUTPUT "Inside Swap: A = ", A, " B = ", B
ENDPROCEDURE

// Main program
DECLARE X : INTEGER
DECLARE Y : INTEGER
X ← 5
Y ← 9
CALL Swap(X, Y)
OUTPUT "After call: X = ", X, " Y = ", Y

(a) Copy and complete a trace table showing the value of A, B and Temp after each of the three assignment statements inside Swap, and state the values output by the OUTPUT statement inside Swap. [3]

(b) State the values output by the final line, OUTPUT "After call: X = ", X, " Y = ", Y, and explain, in terms of how a BYVAL parameter works, why these values are not the swapped values seen inside Swap. [2]

(c) State the one change needed to the header line of Swap so that calling CALL Swap(X, Y) actually swaps the values stored in X and Y in the main program, and explain why this change achieves a real swap. [2]

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Worked solution

Part (a): Tracing the BYVAL call

Swap(X, Y) is called with X = 5 and Y = 9. Because both parameters are declared BYVAL, A and B are separate local variables that each receive a copy of the value passed in: A starts as 5, B starts as 9.

StatementABTemp
On entry to Swap (copies of X and Y)59-
Temp ← A595
A ← B995
B ← Temp955

After the three assignments, A = 9 and B = 5, so the OUTPUT statement inside Swap displays A = 9, B = 5. [3 marks]: [1] for the correct value of Temp (5), [1] for the correct final values of A (9) and B (5), [1] for correctly stating the values output inside Swap.

Part (b): Why X and Y are unaffected

The final line outputs X = 5 and Y = 9, exactly the values X and Y held before the call, not the swapped values seen inside Swap. [1 mark]

This is because BYVAL copies the value of the argument into the parameter when the procedure is called. A and B are entirely separate variables from X and Y, local to Swap, so changing A and B inside the procedure only changes those local copies. Once Swap finishes running, A, B and Temp cease to exist, and X and Y in the main program were never touched. [1 mark] for correctly explaining that BYVAL operates on a copy, so the caller’s variables cannot be modified by the procedure.

Part (c): Fixing Swap with BYREF

Changing the header to:

PROCEDURE Swap(BYREF A : INTEGER, BYREF B : INTEGER)

is the one change needed. [1 mark]

With BYREF, A and B are not copies, they are direct references to whichever variables are supplied as arguments in the call, so inside Swap, A is X and B is Y for the duration of the call. Any assignment to A or B is therefore an assignment to X or Y themselves. Retracing the call with this change: Temp ← A (Temp = 5), A ← B (so X becomes 9), B ← Temp (so Y becomes 5), giving After call: X = 9, Y = 5, a genuine swap. [1 mark] for explaining that BYREF parameters are aliases of the caller’s variables rather than copies.

Final answers

  • (a) A = 9, B = 5, Temp = 5; Swap outputs A = 9, B = 5.
  • (b) X = 5, Y = 9 (unchanged), because BYVAL parameters are copies local to Swap.
  • (c) Change both parameters to BYREF, since BYREF parameters alias the caller’s variables directly, so the call would then output X = 9, Y = 5.