String handling operations in a programming language

Length

Returns the number of characters in a string.

LEN('Hello')        # 5
LEN('AQA 8525')     # 8  — the space counts
LEN('')             # 0  — an empty string
LEN(' ')            # 1  — a single space is one character

Two facts that carry marks:

• A space is a character. 'AQA 8525' is 8 characters, not 7.
• An empty string has length 0. This is exactly the specification's "checking if a string is empty" validation routine (3.2.11).

Where length is actually used

# Minimum length check — a named AQA validation routine
IF LEN(password) < 8 THEN
    OUTPUT 'Password must be at least 8 characters'
ENDIF

# Empty check — the other named routine
IF LEN(name) = 0 THEN
    OUTPUT 'You must enter a name'
ENDIF

# Loop over every character
FOR i ← 0 TO LEN(word) - 1
    OUTPUT word[i]
ENDFOR

Note LEN(word) - 1 — the same off-by-one as arrays, for the same reason.

Position

Returns where a character or substring first occurs in a string.

POSITION('Computer', 'p')      # 3  (zero-indexed: C=0, o=1, m=2, p=3)

⚠️ Index base. Positions are normally counted from 0, so the first character is at position 0. Some questions count from 1 — the question will tell you, so read carefully before answering.

What if it is not there? Most languages return a value such as −1 to indicate "not found". You are not expected to memorise a specific return value, but you should recognise that a search which fails needs a way to report it.

AQA tip. Length questions look trivial and are frequently got wrong by miscounting a space or by forgetting that an empty string is still a valid string. Count characters deliberately, marking spaces.

Substring extracts a section of a string.

SUBSTRING('Computer Science', 0, 7)     # 'Computer'

Conventions vary between languages — some take (start, length), others (start, end). The specification simply requires that you can use substring, so an exam question will always make its convention clear, either by stating it or by showing a worked example. Read that carefully before answering.

Common uses

# First character (e.g. an initial)
initial ← SUBSTRING(firstName, 0, 1)

# Extract the year from a date string 'DD/MM/YYYY'
year ← SUBSTRING(dateOfBirth, 6, 4)

# Extract the area code from a phone number
areaCode ← SUBSTRING(phone, 0, 5)

# Split a postcode
outward ← SUBSTRING(postcode, 0, 4)

Building a username — the classic exam task

"Create a username from the first 3 letters of the surname plus the first initial."

surname   ← 'Patel'
firstName ← 'Amina'

username ← SUBSTRING(surname, 0, 3) + SUBSTRING(firstName, 0, 1)
# 'Pat' + 'A' = 'PatA'

This single example uses substring and concatenation together, which is exactly how the subtopic is examined.

Guarding against a short string

IF LEN(surname) ≥ 3 THEN
    code ← SUBSTRING(surname, 0, 3)
ELSE
    code ← surname
ENDIF

Asking for three characters from a two-character surname causes an error, so a robust program checks the length first. Noticing that is a genuine AO3 point.

AQA tip. When a question gives a substring example, use its convention exactly — do not substitute the one from your taught language.

Concatenation joins two or more strings together end to end. In AQA pseudo-code the operator is +.

firstName ← 'Amina'
surname   ← 'Patel'

fullName ← firstName + ' ' + surname     # 'Amina Patel'

Note the ' '. Concatenation joins strings exactly — it adds no spaces. Without the explicit space you get 'AminaPatel'. Forgetting it is a very common slip.

The + overload — a real source of confusion

5 + 3          = 8       # numbers → addition
'5' + '3'      = '53'    # strings → concatenation
5 + '3'        = ERROR   # mixed → usually an error

The same symbol does two completely different jobs depending on the data types. This is why '5' + '3' gives '53' rather than 8 — a classic exam question, and the reason input must be converted before arithmetic (3.2.1).

Building output messages

OUTPUT 'Hello ' + name + ', you scored ' + INT_TO_STRING(score) + ' points'

The number must be converted to a string before it can be concatenated onto text — concatenation joins strings, and a number is not a string.

AQA tip. In "build this string" questions, write out the intermediate result at each step. 'Pat' + 'A' = 'PatA' shown explicitly is worth more than a bare final answer if you slip later.

Every character is stored as a number — its character code (3.3.5). Two operations convert between them.

CHAR_TO_CODE('A')      # 65   — character → code
CODE_TO_CHAR(66)       # 'B'  — code → character

The ASCII values worth knowing

Characters	Codes
'A' to 'Z'	65 to 90
'a' to 'z'	97 to 122
'0' to '9'	48 to 57
space	32

You do not need to memorise a full table — questions supply one when needed. But you must know that codes run in sequence, so once you know one you can calculate the rest:

• 'A' = 65, so 'B' = 66, 'C' = 67, and 'E' = 65 + 4 = 69.
• 'a' = 97, so 'e' = 97 + 4 = 101.

The uppercase/lowercase gap

$\text{lowercase code} - \text{uppercase code} = 32$

'a' (97) − 'A' (65) = 32, and this holds for every letter. So:

# Convert an uppercase letter to lowercase
lower ← CODE_TO_CHAR(CHAR_TO_CODE(letter) + 32)

# Convert lowercase to uppercase
upper ← CODE_TO_CHAR(CHAR_TO_CODE(letter) - 32)

This is also why string comparison is case-sensitive (3.2.4): 'A' and 'a' have different codes, so 'YES' = 'yes' is False.

The digit trap

CHAR_TO_CODE('7')      # 55, NOT 7

The character '7' has code 55; the number 7 is a different thing entirely. To get the numeric value of a digit character you can subtract the code of '0':

value ← CHAR_TO_CODE('7') - CHAR_TO_CODE('0')     # 55 - 48 = 7

AQA tip. When a question gives you a character encoding table, use it rather than relying on memory — and remember the sequence rule, which lets you work out any code in a group from a single given value.

The specification lists exactly four expected string conversion operations:

Conversion	Purpose	Example
string → integer	Turn typed input into a whole number	STRING_TO_INT('25') → 25
string → real	Turn typed input into a decimal number	STRING_TO_REAL('19.99') → 19.99
integer → string	Join a whole number onto text	INT_TO_STRING(25) → '25'
real → string	Join a decimal number onto text	REAL_TO_STRING(19.99) → '19.99'

Why they exist — the two boundaries of a program

# ON THE WAY IN — keyboard input is a string
age ← STRING_TO_INT(USERINPUT)
nextYear ← age + 1                    # now the arithmetic works

# ON THE WAY OUT — concatenation joins strings
OUTPUT 'Next year you will be ' + INT_TO_STRING(nextYear)

Conversion happens at both ends: coming in from the keyboard, and going out to the display. The one students forget is the second.

Choosing between INT and REAL

Ask whether a decimal part is needed:

Value	Conversion	Why
Number of tickets	STRING_TO_INT	A count is always whole
Price in pounds	STRING_TO_REAL	Needs pence
A person's age in years	STRING_TO_INT	Whole years
A weight in kg	STRING_TO_REAL	Could be 3.25
A percentage result	STRING_TO_REAL	Could be 74.5

Using STRING_TO_INT on '19.99' either fails or truncates to 19, losing the pence.

The robustness caveat

age ← STRING_TO_INT(USERINPUT)      # crashes if the user types 'twenty'

Converting a non-numeric string causes an error. A robust program checks the input is numeric before converting — a link forward to 3.2.11.

AQA tip. Marks for conversion are awarded in questions that are ostensibly about something else — validation, calculation, output formatting. Making the three-line unit (prompt → input → convert) automatic protects marks right across Paper 1.

Exam questions rarely use one operation alone. Three patterns recur.

Pattern 1 — build a username or code

# First 3 of surname + first initial + last 2 digits of year
username ← SUBSTRING(surname, 0, 3)
         + SUBSTRING(firstName, 0, 1)
         + SUBSTRING(year, 2, 2)

# 'Patel', 'Amina', '2026'  →  'Pat' + 'A' + '26'  =  'PatA26'

Uses substring ×3 and concatenation ×2.

Pattern 2 — validate then convert

REPEAT
    OUTPUT 'Enter a mark between 0 and 100'
    input ← USERINPUT

    IF LEN(input) = 0 THEN
        OUTPUT 'You must enter something'
    ENDIF
UNTIL LEN(input) > 0

mark ← STRING_TO_INT(input)

Uses length for validation and string→integer for conversion.

Pattern 3 — process every character

# Count the vowels in a word
vowelCount ← 0

FOR i ← 0 TO LEN(word) - 1
    letter ← SUBSTRING(word, i, 1)

    IF letter = 'a' OR letter = 'e' OR letter = 'i'
       OR letter = 'o' OR letter = 'u' THEN
        vowelCount ← vowelCount + 1
    ENDIF
ENDFOR

Uses length for the loop bound and substring to isolate each character. Note each side of the OR is a complete condition (3.2.5).

The method for chained-operation questions

1. Write the starting string.
2. Apply one operation and write the result.
3. Repeat, writing every intermediate value.
4. State the final answer.

Showing intermediates earns method marks even if a later step slips.

AQA tip. With nested operations such as CODE_TO_CHAR(CHAR_TO_CODE('A') + 2), work from the inside out: CHAR_TO_CODE('A') = 65, then 65 + 2 = 67, then CODE_TO_CHAR(67) = 'C'.