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Data Types questions

  1. ​

    1. Describe why two's complement may be preferable to sign and magnitude
      • Two's complement allows addition to work with negative numbers as if the numbers were positive, and subtraction to work by inverting the bits of the right number, adding the result to the left number, and adding one. Additionally, two's complement does not allow a representation of -0, meaning it gives an extra number, whereas sign and magnitude provides -0 and therefore has one fewer unique number
    2. Demonstrate subtraction in binary using 8-bit two's complement using the equivalent of the denary calculation 47-23. You must show all working.

      • 47 = 0x30 - 1 = 0x2F = 0b00101111
        -23 = ~22 = ~0x16 = 0xE9 = 0b11101001
        47 - 23 = 47 + -23

        0b00101111
        0b11101001
        b111011110
        0b00011000

        = 0x18 = 24

  2. ​

    1. ​
      1. Change the denary number -89 into a two's complement, 8-bit binary number.
        • -89 = ~88 = ~0x58 = 0xA7 = 0b10100111
      2. Change the denary number -72 into a two's complement, 8-bit binary number.
        • -72 = ~71 = ~0x47 = 0xB8 = 0b10111000
    2. ​

      1. Add the two binary answers which you obtained, using 8-bit arithmetic.

        You must show your working.

        • 0b10100111
          0b10111000
          b101000000
          0b01011111

          = 0x5F = 95

      2. Explain why your answer to the addition sum is wrong

        • When adding the two negative numbers, integer overflow occurs and therefore the sum is off by 256 - the correct answer to the addition is 95 - 256 = -161

  3. Express the denary number -43 in binary using 8-bit two's complement representation.

    Show your working.

    • -43 = ~42 = ~0x2A = 0xD5 = 0b11010101

  4. Show a representation of denary -119 in 8-bits using:

    1. Sign and Magnitude
      • 0b11110111
    2. Two's Complement
      • 0b10001001

  5. Convert the denary number -19 to an 8-bit number using:

    1. Two's complement representation.
      • 0b11101101
    2. Sign and Magnitude representation.
      • 0b10010011

  6. Using two’s complement convert the denary number −43 into an 8 bit binary number. You must show your working.

    • -43 = ~42 = ~0x2A = 0xD5 = 0b11010101

  7. Convert the denary number -52 into an 8-bit binary number using two’s complement.

    • -52 = ~51 = ~0x33 = 0xCC = 0b11001100

  8. Convert the denary number -8 to:

    1. An 8-bit sign and magnitude binary number.
      • 0b10001000
    2. An 8-bit two's complement binary number.
      • 0b11111000

  9. ​

    1. ​
      1. Convert the denary number 188 to an unsigned 8-bit binary number.
        • 0xBC = 0b10111100
      2. Convert the denary number 188 to hexadecimal.
        • 0xBC
    2. ​
      1. Convert the denary number -44 to an 8-bit binary number with sign and magnitude representation.
        • 0b10101100
      2. Convert the denary number -44 to an 8-bit binary number with two's complement representation.
        • 0b11010100

  10. ​

    1. Variables in programs contain specific types of data.

      Complete the table below to suggest a suitable data type for each piece of data.

Data Data Type
'H' Character
"Hello" String
35 Integer
-2.625 Real
True Boolean 
02. Show the denary number 35 as an 8-bit unsigned binary number.
    - 0b00100011u
03. The character 'A' in the ASCII character set is represented by the denary value 65. Write the binary representation for the ASCII character 'H'. Show your working.
    - 'H' = 'A' - 1 + 8 = 'A' + 7 = 72

  1. ​

    1. A simple program is shown below.

      //Program to calculate number of times
//a number goes into 100
count = 0
num = int(input("Enter a number"))
while (count*num)<=100
    count=count+1
endwhile
count=count-1 //Take one off as gone over

print(str(num) + " goes into 100 " + str(count) + " times.")

      State the most suitable data type of the variable count

      • int

    2. State the data type of the result of the expression (count*num)<=100

      • Boolean

    3. State the data type of the result of the expression str(num) + " goes into 100 " + str(count) + " times."

      • String

  2. Convert the unsigned binary number 0b11110000u to:

    1. Denary:
      • 240
    2. Hexadecimal:
      • 0xF0

  3. Below are extracts from the ASCII and EBCDIC character sets.

    ASCII

Denary value Character
65 A
66 B
67 C
68 D
69 E
70 F
71 G
72 H
73 I
74 J
75 K
76 L
77 M
78 N
79 O
80 P
81 Q
82 R
83 S
84 T
85 U
86 V
87 W
88 X
89 Y
90 Z 
EBCDIC
Denary value Character
193 A
194 B
195 C
196 D
197 E
198 F
199 G
200 H
... ...
209 J
210 K
211 L
212 M
213 N
214 O
215 P
216 Q
217 R
... ...
226 S
227 T
228 U
229 V
230 W
231 X
232 Y
233 Z 
Explain, referring to ASCII and EBCDIC, what would happen if computers were to use different character sets when communicating.
- If computers were to use different character sets when communicating, they would misunderstand each other. If computer A, using ASCII, attempts to transmit `Z` to computer B, using EBCDIC, it will arrive as a completely different character (character 90 EBCDIC, 103 characters before `A` in EBCDIC)
  1. ​
    1. Convert the denary number 72 to an unsigned 8-bit integer.
      • 0b01001000u
    2. Convert the unsigned binary number 0b10000101u to denary.
      • 165
    3. Convert the denary number 104 to hexadecimal.
      • 0x68
    4. Given that computers store everything in binary, explain how they are able to represent text.
      • Computers are able to store text as a sequence of bytes, which can then be decoded with a character set such as UTF-8, UTF-16, EBCDIC, or ASCII
  2. ​

    1. Demonstrate how the bytes below are added together. Show your working.

      0b01101010
      0b00111111
      0b11111100
      0b10101001

    2. Convert the binary number shown below to hexadecimal.

      • 0x370F
  3. ​
    1. Show how the binary number 0b01011110 is represented in hexadecimal.
      • 0x5E
    2. Show how the hexadecimal number 0x9B is represented in denary.
      • 9 × 16 + 11 = 144 + 11 = 155
    3. Show how the denary number -87 is represented in sign and magnitude binary.
      • 0b10000000 | 0b1010111 = 0b11010111