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Data representation

A-Level Computer Science 9618 · Topic 13 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. User-defined data types
  2. File organisation, access and hashing
  3. Floating-point binary representation
  4. Normalisation, range and precision

1User-defined data types

Syllabus 13.1

A user-defined data type is one the programmer defines, derived from the built-in types, because the built-in types alone do not model the data in the problem. A non-composite type is defined without reference to other user types: an enumerated type lists every value it can take, in order, and a pointer type holds the memory address of a value of a stated type. A composite type groups other data types under one identifier: a record (fields of different types), a set (unordered, no duplicates, supports set operations such as membership) and a class, whose objects combine attributes with methods.

User-defined data types Non-composite Composite defined without other types built from other data types EnumeratedPointer RecordSetClass / object (Spring, Summer, ...)^INTEGER TYPE ... ENDTYPESET OF CHAR the programmer defines the type when the built-in types do not fit the problem
Fig. 1: the two families of user-defined data type named in the 9618 syllabus

Worked example: Trace a pointer

Using the code above, what are the values of Score and MyPointer^ after the last line? What would OUTPUT MyPointer display?
  1. Score ← 40 stores 40 in the memory location called Score.
  2. MyPointer ← ^Score stores the address of Score in MyPointer, not the value 40.
  3. MyPointer^ ← MyPointer^ + 2 reads the value at that address (40), adds 2 and writes 42 back to the same location, which is Score.
  4. So Score and MyPointer^ are both 42. OUTPUT MyPointer would display a memory address, not 42.
Answer: Score = 42 and MyPointer^ = 42; MyPointer holds the address of Score

Shape, then syntax.

  • Non-composite: enumerated TYPE T = (A, B, C); pointer TYPE P = ^INTEGER.
  • Pointers: ^X is the address of X; Ptr^ is the value at the address in Ptr.
  • Composite: record (TYPE ... ENDTYPE), set (SET OF + DEFINE), class (CLASS ... ENDCLASS, NEW).
  • Choose by shape: fixed list of values, address, mixed fields, membership only, data plus methods.

Watch out!

Writing TYPE Season = ("Spring", "Summer", ...) loses the mark. Enumerated values are not strings, so they have no quotation marks.

A set is not an array: it has no index and no order, and adding a value that is already present changes nothing. Do not describe a set as "a list of values with positions".

Revisit it: User-defined data types lesson · worked practice

2File organisation, access and hashing

Syllabus 13.2

File organisation is how records are arranged in a file. In a serial file records are stored in the order they arrive; in a sequential file they are stored in order of a key field; in a random file each record is stored at an address calculated from its key by a hashing algorithm. File access is how a record is reached: sequential access reads records one after another from the start (serial and sequential files), while direct access goes straight to the record (random files via the hash, sequential files via an index). When two keys hash to the same address there is a collision, handled by storing the record in the next free address or in an overflow area.

SerialSequentialRandom order of arrivalin key orderaddress = hash(key) 317102455208131 102131208317455 both read from the start, one record after another 208 0123456 208 MOD 7 = 5 a hash on the key gives the address, so the record is reached directly
Fig. 1: the same five records in a serial, a sequential and a random file

Worked example: Choose the organisation

An electricity company produces a bill for every customer once a month. Its call centre must also show any single customer's account within a second. Which organisation and access method suits each task?
  1. Monthly billing processes almost every record (a high hit rate), in key order, in one batch run.
  2. So a sequential file accessed sequentially suits billing: no time is wasted calculating addresses.
  3. The call centre needs one record at a time, quickly, with a low hit rate per run.
  4. So direct access suits it: a random file with a hash on the account number (or an indexed sequential file).
Answer: Billing: sequential file, sequential access. Look-up: random file (or indexed sequential), direct access.

Organise, then access.

  • Serial: order of arrival. Sequential: key order. Random: address = hash(key).
  • Sequential access reads from the start; direct access jumps to the record.
  • High hit rate, batch: sequential. One record at a time, fast: random.
  • Collisions: next free address (wrap round) or an overflow area; always check the key you read.

Watch out!

Serial and sequential are not the same. A serial file has no order; a sequential file is sorted on a key field. Say which key the file is ordered by to earn the mark.

After hashing, always compare the key of the record found with the key wanted. The record at the home address may belong to a different key that collided there.

Revisit it: File organisation, access and hashing lesson · worked practice

3Floating-point binary representation

Syllabus 13.3

A floating-point number is stored as mantissa × 2exponent. The mantissa holds the significant digits and the exponent says how many places the binary point moves. In 9618 both are in two's complement, and the binary point sits just after the first bit of the mantissa, so the mantissa's place values are −1, ½, ¼, 1/8 and so on. A positive exponent moves the point to the right (a bigger number); a negative exponent moves it to the left (a smaller number). With the same number of bits, floating point gives a far wider range than fixed point.

8-bit mantissa and 4-bit exponent, both in two’s complement −1½¼1/81/161/321/641/128 −8421 01011000 0011 binary point Mantissa Exponent Mantissa = ½ + 1/8 + 1/16 = 0.6875 Exponent = 2 + 1 = 3 Value = 0.6875 × 2³ = 5.5 or move the point 3 places right: 0.1011 → 101.1 = 5.5
Fig. 1: an 8-bit mantissa and 4-bit exponent representing 5.5

Worked example: Negative mantissa

Convert the floating-point number 10100000 0010 (8-bit mantissa, 4-bit exponent, both two's complement) to denary.
  1. Exponent 0010 = +2.
  2. Mantissa 1.0100000: place values −1 and ¼ are set, so the mantissa is −1 + 0.25 = −0.75.
  3. Value = −0.75 × 22 = −0.75 × 4.
  4. Check by moving the point 2 places right: 101.00000, and in two's complement 101 is −4 + 1 = −3.
Answer: −3

Exponent first.

  • Value = mantissa × 2exponent; both in two's complement.
  • Mantissa place values: −1 . ½ ¼ 1/8 ...; the point follows the first bit.
  • To denary: exponent, mantissa, multiply (or move the point); show each step.
  • From denary: binary, slide the point to 0.1, count places for the exponent; negate the mantissa for negatives and check it starts 10.

Watch out!

Do not treat the mantissa as a whole number. 01011000 is not 88: the point sits after the first bit, so it is 0.1011000 = 0.6875.

Negating a mantissa can leave it starting 11. For −4: 4 = 0.1 × 23, so 01000000 0011 negates to 11000000 0011, which is not normalised. Shift left once and reduce the exponent: the correct answer is 10000000 0010.

Revisit it: Floating-point binary representation lesson · worked practice

4Normalisation, range and precision

Syllabus 13.3

A floating-point number is normalised when the first two bits of the mantissa are different: a positive mantissa starts 0.1 and a negative one starts 1.0. To normalise, shift the mantissa left until this is true and subtract the number of places shifted from the exponent, so the value is unchanged. Normalisation gives the maximum precision for the number of mantissa bits and a unique representation for each number. For a fixed total, more mantissa bits give greater precision and more exponent bits give a greater range. Many denary fractions, such as 0.1, recur in binary and can only be stored as an approximation, causing rounding errors; a result too large for the exponent causes overflow, and one too close to zero causes underflow.

normalising: same value, more significant bits kept beforeafter 0.00110100.1101000 01010011 E = 5E = 3 mantissa: 2 places left exponent − 2 both are 6.5, but now the first two bits differ positive normalised: starts 0.1 negative normalised: starts 1.0
Fig. 1: normalising 00011010 0101 gives 01101000 0011; both represent 6.5

Worked example: Positive mantissa

Normalise 00011010 0101 (8-bit mantissa, 4-bit exponent, both two's complement).
  1. The mantissa starts 00, so it is not normalised.
  2. Shift left until it starts 01: 00011010 → 00110100 → 01101000, which is 2 places.
  3. Exponent 0101 = 5; subtract 2 to get 3 = 0011.
  4. Check: before, 0.0011010 = 0.203125 × 25 = 6.5; after, 0.1101 = 0.8125 × 23 = 6.5.
Answer: 01101000 0011

Two different bits.

  • Normalised: positive starts 01, negative starts 10.
  • Normalise: shift the mantissa left, subtract the shift count from the exponent; check the value is unchanged.
  • More mantissa bits: precision up. More exponent bits: range up.
  • Limits: recurring binary fractions give rounding errors; too big is overflow, too close to zero is underflow.

Watch out!

Shifting the mantissa without changing the exponent changes the number. Every left shift must take 1 off the exponent; state the new exponent in binary.

Overflow and underflow are about the size of the result, not about a negative number. Underflow means too close to zero, not "too negative".

Revisit it: Normalisation, range and precision lesson · worked practice
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