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A-Level Computer Science 9618 · Topic 11 revision notes

Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

  1. Variables, constants and built-in functions
  2. Selection and iteration constructs
  3. Procedures and functions
  4. Passing parameters by value and by reference

1Variables, constants and built-in functions

Syllabus 11.1

A variable is a named memory location whose value can change while the program runs; declare it with DECLARE Name : TYPE and give it a value with the assignment operator ←. A constant is declared once with CONSTANT Name = value, using a literal only, and can never be changed. Expressions combine values with arithmetic operators (+ - * / DIV MOD), relational operators and the logical operators AND, OR, NOT. Built-in functions and library routines are pre-written, pre-tested subroutines such as LENGTH, MID and INT.

each flowchart symbol becomes one pseudocode statement STARTINPUT Celsius F ← Celsius * 9 / 5 + 32OUTPUT FSTOP DECLARE Celsius : REAL DECLARE F : REAL INPUT Celsius F ← Celsius * 9 / 5 + 32 OUTPUT F START and STOP need no statement; declarations come first Celsius = 20 gives F = 68
Fig. 1: converting a program flowchart into pseudocode, one symbol at a time

Worked example: Structured English to pseudocode

Convert this design into pseudocode. 1. SET the hourly rate to 12.50 (it never changes). 2. INPUT the hours worked. 3. CALCULATE the pay as hours multiplied by the rate. 4. OUTPUT the pay. Then state the output when 6 hours are entered.
  1. Step 1 is a value that never changes, so it is a constant: CONSTANT HourlyRate = 12.50.
  2. Declare the variables with suitable types: DECLARE Hours : INTEGER and DECLARE Pay : REAL.
  3. Steps 2 and 3: INPUT Hours then Pay ← Hours * HourlyRate.
  4. Step 4: OUTPUT "Pay: ", Pay. With 6 hours, 6 × 12.50 = 75.00.
Answer: 
CONSTANT HourlyRate = 12.50
DECLARE Hours : INTEGER
DECLARE Pay : REAL
INPUT Hours
Pay ← Hours * HourlyRate
OUTPUT "Pay: ", Pay
With 6 hours the output is Pay: 75.00 (many languages show it as 75.0).

Literal, operator, function.

  • Constant: CONSTANT Name = literal, declared at the start, never assigned again.
  • Division: / gives a REAL; DIV gives the whole-number quotient; MOD gives the remainder.
  • Strings: positions start at 1; MID(s, start, length); join with &.
  • Random integers: INT(RAND(n)) gives 0 to n − 1; add the lowest value you want.

Watch out!

CONSTANT VAT = Price * 0.2 is not allowed: a constant must be a literal. And a constant can never appear on the left of ← later in the program.

17 / 5 is 3.4, not 3. If a question says the result must be an INTEGER, the mark is for DIV (or INT), not /.

In the 9618 guide UCASE and LCASE take a single CHAR, and string positions start at 1. To change a whole string, loop through it with MID one character at a time.

Revisit it: Variables, constants and built-in functions lesson · worked practice

2Selection and iteration constructs

Syllabus 11.2

Selection uses IF ... THEN ... ELSE ... ENDIF (which can be nested) or CASE OF ... OTHERWISE ... ENDCASE, whose clauses are tested in order until one matches. Iteration has three forms: a count-controlled loop (FOR ... NEXT) repeats a known number of times; a post-condition loop (REPEAT ... UNTIL) tests at the end, so it always runs at least once; a pre-condition loop (WHILE ... ENDWHILE) tests at the start, so it may run zero times.

FOR (count)WHILE (pre)REPEAT (post) i ← 1i ≤ 10? bodyi ← i + 1 cond?body bodycond? TRUEFALSE TRUEFALSE FALSETRUE runs a fixed number of timesmay run 0 timesruns at least once
Fig. 1: where each loop tests its condition decides the fewest times its body can run

Worked example: Dry-run the delivery charge

Give the output for (a) Weight 14.0, Express FALSE and (b) Weight 2.0, Express TRUE.
  1. (a) 14.0 <= 2.0 is FALSE, so the ELSE runs; 14.0 <= 10.0 is FALSE, so the inner ELSE runs.
  2. Charge ← 7.50 + (14.0 − 10.0) × 0.50 = 7.50 + 2.00 = 9.50. Express is FALSE, so no change.
  3. (b) 2.0 <= 2.0 is TRUE (it is a boundary value), so Charge ← 4.00.
  4. Express is TRUE, so Charge ← 4.00 × 2 = 8.00.
Answer: (a) Delivery charge: 9.50 (b) Delivery charge: 8.00

Test first or test last.

  • IF: THEN on the IF line, one ENDIF per IF; check the boundary values.
  • CASE: one variable, values or ranges, first match wins, OTHERWISE last.
  • FOR = known count; REPEAT = at least once; WHILE = maybe zero times.
  • Justify a loop by saying why its test position suits this problem, not just what the loop is.

Watch out!

Check boundary values. Changing <= to < moves a 2.0 kg parcel into the next price band, and examiners test exactly that value.

CASE stops at the first match. Overlapping ranges are not an error, but every value is caught by the earlier range, so the later clause is dead code.

The UNTIL condition is the condition to stop; the WHILE condition is the condition to keep going. Converting one loop into the other means reversing the condition.

Revisit it: Selection and iteration constructs lesson · worked practice

3Procedures and functions

Syllabus 11.3

A procedure is a named block of code that performs a task; it may have none, one or more parameters and is run with a CALL statement. A function also performs a task but returns a single value: its header ends RETURNS data type, it uses RETURN, and it is used inside an expression, where the return value replaces the call. The values passed in a call are arguments; the identifiers in the header that receive them are parameters. Subroutines give modular, reusable code that is easier to test and maintain.

a procedure call is a statement; a function call is part of an expression main program Total ← 10 CALL ShowLine(5, '*') Total ← Total + Square(4) OUTPUT Total PROCEDURE ShowLine(...)builds and outputs *****ENDPROCEDURE FUNCTION Square(N ...)RETURN N * NENDFUNCTION returns control only 16 replaces the call: Total = 26
Fig. 1: control passes to the subroutine and back; only a function brings a value back with it

Worked example: Follow the calls

State exactly what the last two lines of the program above output.
  1. CALL ShowBanner() runs ShowBanner, which first calls ShowLine with Length 20 and Symbol '='.
  2. ShowLine joins '=' to Line 20 times and outputs a row of 20 = signs.
  3. ShowBanner then outputs the title and calls ShowLine(20, '=') again.
  4. Control returns to the main program: CALL ShowLine(5, '*') outputs *****.
Answer: A row of 20 = signs, the line Weekly quiz results, another row of 20 = signs, then *****

Statement or expression.

  • Procedure: PROCEDURE Name(...) ... ENDPROCEDURE; run with CALL Name(...).
  • Function: header ends RETURNS type; uses RETURN; used inside an expression, never with CALL.
  • Parameter is in the header; argument is in the call; they match in number, order and type.
  • Efficient: stop loops early, do not repeat calculations, replace copied code with a subroutine.

Watch out!

A procedure does not return a value, so X ← ShowLine(5, '*') is wrong. If the main program needs a result back, write a function instead.

CALL CountChar("banana", 'a') loses the mark: a function is never CALLed, and its return value would be thrown away.

Do not mix up parameter and argument. The parameter is the name in the header (Length); the argument is what is passed in the call (4.0).

Revisit it: Procedures and functions lesson · worked practice

4Passing parameters by value and by reference

Syllabus 11.3

Passing by value (BYVAL) gives the subroutine a copy of the argument's value: changes to the parameter are local and the caller's variable is unchanged. Passing by reference (BYREF) passes the address of the caller's variable, so the parameter is another name for the same memory location and any change is made to the original. If neither keyword is given, passing by value is assumed. In 9618 pseudocode, parameters should not be passed by reference to a function.

BYVAL: a copy BYREF: the original callerprocedurecallerprocedure X = 5 A = 5 value copied A ← A * 2 A becomes 10, X stays 5 Y = 3 B address passed B ← B + 10 B is another name for Y: Y becomes 13
Fig. 1: BYVAL copies the value into a new location; BYREF lets the parameter refer to the caller's own variable

Worked example: Trace a BYVAL call

Trace the program above and give both lines of output.
  1. Score ← 7 in the main program.
  2. CALL Double(Score): the value 7 is copied into the parameter N.
  3. N ← 7 × 2 = 14 and "Inside: 14" is output. Only the copy changed.
  4. The procedure ends, N is discarded, and "Outside: 7" is output.
Answer:
ScoreNOUTPUT
7
7
14Inside: 14
7Outside: 7

Value in, reference out.

  • BYVAL: a copy; changes stay inside the subroutine; the default if nothing is stated.
  • BYREF: the address; the parameter is the caller's variable; the argument must be a variable.
  • One keyword applies to all the following parameters until the other keyword appears.
  • Functions take BYVAL parameters only and return one value; use BYREF in a procedure to pass back several values.

Watch out!

"The procedure changes N, so Score changes" is the classic wrong answer. With BYVAL the procedure only ever touches its own copy.

In PROCEDURE Swap(BYREF X : INTEGER, Y : INTEGER), Y is passed by reference too. Saying Y is passed by value loses the mark.

If a question says a procedure must "return" two values, the answer is BYREF parameters, not two RETURN statements: a procedure has no RETURN value, and a function returns only one value.

Revisit it: Passing parameters by value and by reference lesson · worked practice
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