13 - data representation

Question 1

user defined data type

Answer 1

a data type based on an existing data type or other data types that have been defined by a programmer

Question 2

why use user defined data types

Answer 2

• built in types arent sufficient to represent data
• store data of same or diff types (flexible)
• can be encapsulated which ensures consistency and uniformity
• Can be reused

Question 3

non-composite data type

Answer 3

a data type that doesnt reference other data types
eg enumerated, pointer

Question 4

enumerated data type

Answer 4

non composite data type
defined by a list of all possible values with an implied order
eg TSeasons = (Spring ,Summer, Autumn, Winter)
DECLARE thisSeason : TSeasons
TYPE <identifier> = (value1, value2, value3...)</identifier>

Question 5

pointer data type

Answer 5

non composite
references a memory location
TYPE <pointer> = ^<Typename>
eg TYPE TmonthPointer = ^Tmonth
DECLARE monthPointer : TmonthPointer</Typename></pointer>

Question 6

set data type

Answer 6

composite
- a list of unordered elements that can use set theory operations eg intersection, union, difference
- Contains values that are not organised
- Cant have duplicate values
TYPE <set> = SET OF <basetype>
DEFINE <identifier> (value1, value2...) : <set>
eg TYPE Sletter = SET OF CHAR
DEFINE vowel ('a', 'e', 'i', 'o', 'u') : letters</set></identifier></basetype></set>

Question 7

composite data type

Answer 7

uses multiple built in data types
eg record, class, set

Question 8

record data type

Answer 8

TYPE <idenfiier>
DECLARE <identifier> : STRING
....</identifier></idenfiier>

Question 9

class data type

Answer 9

composite
includes variables of given data types and methods
an object is defined from a class

Question 10

serial files

Answer 10

records of data are stored in a file, one after another, in the order they were added to the file
- new records appended to the end
eg meter readings, customer transactions

Question 11

sequential files

Answer 11

records of data are stored in a file, one after another, in a given order (defined by the key field)
eg store all bank transactions

Question 12

random files

Answer 12

records of data are stored in a file in any available position
the location of any record in the file is found by using a hashing algorithm on the key field of a record

Question 13

hashing algorithm

Answer 13

mathematical formula used to perform a calculation on the key field of the record;
the result of the calculation gives the address where the record should be found

Question 14

hasing collision

Answer 14

eg hash is to divide by 100
102 and 202 will give the same remainder so there is a collision
deal with it by:
- open hash - using the next free space
- closed hash - using the next free space in a give pre set up overflow area
when reading records if key fields dont match when the hash is calculated then all following records need to be read or the overflow area needs to be searches

Question 15

sequential access

Answer 15

records are searched one after another
from the physical start of the file until the required record is found
- used when every record needs to be processed

Question 16

direct access

Answer 16

a record can be physically found in a file without physically reading other records
- used to find records quickly when an individual record needs to be processed
use hashing or index table

Question 17

file access

Answer 17

the method used to physically find a record in the file.
- sequential
- random

Question 18

mantissa

Answer 18

fractional part of floating point num

Question 19

exponent

Answer 19

the power of 2 that the mantissa is raised to

Question 20

binary floating point number

Answer 20

a binary number in the form M*2^E
M- mantissa
E - exponent

Question 21

overflow

Answer 21

the result of carrying out a calculation which produces a value too large for the computer’s allocated word size

Question 22

underflow

Answer 22

the result of carrying out a calculation which produces a value too small for the computer’s allocated word size.

Question 23

converting to floating point

Answer 23

convert to binary
shift to have 1.0 or 0.1 (mantissa)
exponent = amount shifted by

to convert back take exponent and convert - shift by that amount
then convert mantissa to denary

Question 24

fixed point number

Answer 24

• overall number of bits chosen with a defined number for the whole part + remainder for fractional part
• MSB for sign, next 5 for whole number, last two for fractional part
• Eg 011111.11 = 37.75 - largest possible value
  Eg 000000.01 = 0.01 0.25 - smallest positive value

Question 25

rounding errors

Answer 25

some numbers can only be represented as an approximate value (accuracy of which determined by no. b in mantissa)
eg 0.1 will lead to a rounding error has its impossible to convert
reduce error by increasing size of mantissa
if you add rounded numbers it increases the error

Question 26

normalisation

Answer 26

for a + num mantissa starts with 0.1
for a - num mantissa starts with 1.0

Question 27

accuracy and range of numbers

Answer 27

accuracy is increases by increasing no b for mantissa
range of numbers is increases by increasing no b for exponent
both these increase storage needed

Question 28

max/ min numbers using 8b 8b (when normalised)

Answer 28

max positive
01111111 01111111
min positive
01000000 10000000
min negative
10111111 10000000
max negative
10000000 01111111