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Programming Constructs and Operators: Question 4

Syllabus 8.1

Structured 8 marks

A banking app displays a customer's reference code on screen with only the last two characters visible; every character before that is replaced with an asterisk. This pseudocode algorithm produces that masked display, one character at a time, using a post-condition loop.

01 DECLARE Code : STRING
02 DECLARE Index : INTEGER
03 Code ← "48213"
04 Index ← 1
05 REPEAT
06     IF Index <= LENGTH(Code) - 2 THEN
07         OUTPUT "*"
08     ELSE
09         OUTPUT SUBSTRING(Code, Index, 1)
10     ENDIF
11     Index ← Index + 1
12 UNTIL Index > LENGTH(Code)

(a) State the value of LENGTH(Code) - 2 for Code = "48213", and state the smallest value of Index for which the ELSE branch (line 8–9) is taken. [2]

(b) Complete a trace table showing the value of Index and the character output on each pass of the loop, for Code = "48213". Hence state the complete sequence of characters displayed on screen. [4]

(c) A different reference code, Code = "9", is used instead. State what is displayed on screen when the algorithm executes with this value, showing your working. [2]

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Worked solution

Part (a): Working out the masking boundary

LENGTH(Code) for Code = "48213" is 5 (5 characters). So:

LENGTH(Code) - 2 = 5 - 2 = 3

The IF condition on line 6 is Index <= 3. This is true for Index = 1, 2, 3 (so those positions are masked with *), and becomes false for the first time when Index = 4. This is the smallest value of Index for which the ELSE branch runs and the real character is shown instead. [2 marks]: [1] for LENGTH(Code) - 2 = 3, [1] for identifying Index = 4 as the first ELSE case.

Part (b): Tracing the REPEAT…UNTIL loop

Code = "48213" has characters 4 (position 1), 8 (position 2), 2 (position 3), 1 (position 4), 3 (position 5). The loop body always runs at least once (it is a post-condition loop), and stops only once Index > LENGTH(Code) becomes true, i.e. once Index > 5.

PassIndexIndex <= 3 ?Statement executedOutput
11trueline 7: OUTPUT "*"*
22trueline 7: OUTPUT "*"*
33trueline 7: OUTPUT "*"*
44falseline 9: OUTPUT SUBSTRING(Code,4,1)1
55falseline 9: OUTPUT SUBSTRING(Code,5,1)3

After pass 5, Index becomes 6, and UNTIL Index > LENGTH(Code) checks 6 > 5, which is true, so the loop stops. The five characters output in sequence, *, *, *, 1, 3, appear on screen one after another as:

***13

[4 marks]: [1] for correctly identifying the three masked passes (Index = 1, 2, 3), [1] for correctly extracting 1 at Index = 4 and 3 at Index = 5 with SUBSTRING, [1] for a fully correct trace table, [1] for the correct final displayed sequence ***13.

Part (c): The edge case Code = “9”

LENGTH("9") = 1, so:

LENGTH(Code) - 2 = 1 - 2 = -1

On the very first (and only) pass, Index = 1. The condition Index <= LENGTH(Code) - 2 becomes 1 <= -1, which is false, since -1 is a negative number and 1 is not less than or equal to it. So the ELSE branch runs immediately: OUTPUT SUBSTRING("9", 1, 1), which outputs 9. Index becomes 2, and UNTIL 2 > 1 is true, so the loop stops after just one pass.

The screen displays the single character in full, with no masking at all:

9

This makes sense: with only 1 character in Code, there aren’t even 2 characters left to keep visible, let alone any earlier ones to hide. [2 marks]: [1] for correctly evaluating LENGTH(Code) - 2 as negative, [1] for the correct final output 9.

Final answers

  • (a) LENGTH(Code) - 2 = 3; the ELSE branch first runs at Index = 4.
  • (b) Trace as shown above, the screen displays ***13.
  • (c) The screen displays 9 (no masking occurs).