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Answers

A7.5 Bitwise operations and characters

Data representation · A level · OCR H446 1.4.1, AQA 7517 4.5.5.1, Eduqas A500QS 2.3 · about 30 min

BugBotLab

What this lesson is about

Masks with AND, OR and XOR, logical, arithmetic and circular shifts, and characters as codes in ASCII and Unicode.

Questions 6 marks in all

  1. [1 mark]What is 10110110 AND 00001111?

    Answer: 00000110. AND keeps a bit only where both bytes have a 1, so the mask keeps the low four bits, 0110, and clears the rest.

  2. [1 mark]Which operation and mask toggles (flips) the lowest four bits of a byte and leaves the rest unchanged?

    1. AXOR with 00001111
    2. BAND with 00001111
    3. COR with 00001111
    4. DAND with 11110000
    Answer: A. XOR with 1 flips a bit and XOR with 0 leaves it alone. AND with 0 clears and OR with 1 sets.

  3. [1 mark]Give the result of an arithmetic shift right by one place of the 8-bit two's complement number 11110000.

    Answer: 11111000. An arithmetic shift copies the sign bit in on the left, so -16 becomes -8.

  4. [1 mark]What does this program print?

    print(0b10110110 >> 2, 0b00010110 << 1)
    Answer:
    45 44

    10110110 (182) shifted right 2 is 101101 (45). 00010110 (22) shifted left 1 is 00101100 (44).

  5. [1 mark]The ASCII code for the character 0 is 48. What is the denary ASCII code for the character 7?

    Answer: 55. The digit characters are in order, so 7 is 48 + 7 = 55. That differs from the pure binary value of 7, which is 00000111.

  6. [1 mark]Why was Unicode introduced?

    1. AASCII could not represent the characters of most of the world's languages
    2. BASCII codes were too long
    3. CUnicode uses fewer bits for every character
    4. DASCII could not store digits
    Answer: A. 7-bit ASCII has only 128 codes, enough for English. Unicode gives codes to characters from all writing systems, plus symbols such as emoji.

The task: unpack a colour

colour holds 0xFD20, a 16-bit RGB565 colour: bits 15 to 11 are red, bits 10 to 5 green, bits 4 to 0 blue. Using only shifts (>>, <<) and masks (&, |): 1. Extract the fields into r5 (0 to 31), g6 (0 to 63) and b5 (0 to 31) and print r5=<r5> g6=<g6> b5=<b5>. 2. Widen each to 8 bits (0 to 255): r8 is r5 shifted left 3, ORed with r5 shifted right 2; g8 is g6 shifted left 2, ORed with g6 shifted right 4; b8 is made like r8. Print rgb=<r8>,<g8>,<b8> with no spaces. 3. Light the LED with led(r8, g8, b8). Do not use //, %, bin or format, and do not type the answers.

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

colour = 0xFD20

r5 = 0
g6 = 0
b5 = 0
print("r5=" + str(r5), "g6=" + str(g6), "b5=" + str(b5))

The hint students can ask for: Shift the 16-bit value right until the field you want sits at the bottom, then AND it with a mask of that field's width. To widen a field, shift it left to fill the new width and OR in its own top bits to fill the gap at the bottom.

A solution

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

colour = 0xFD20

r5 = (colour >> 11) & 0x1F
g6 = (colour >> 5) & 0x3F
b5 = colour & 0x1F
print("r5=" + str(r5), "g6=" + str(g6), "b5=" + str(b5))

r8 = (r5 << 3) | (r5 >> 2)
g8 = (g6 << 2) | (g6 >> 4)
b8 = (b5 << 3) | (b5 >> 2)
print("rgb=" + str(r8) + "," + str(g8) + "," + str(b8))
led(r8, g8, b8)

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.