Bits and Bytes in Computer Memory

Computer memory is composed of bits representing 0 or 1, grouped into bytes. Learn about binary operations, converting decimal to binary, octal representation, and exercises in this module. Understand how individual bits in a byte correlate to powers of 2 and perform binary exercises to strengthen your understanding. Explore the conversion process from decimal to binary and the concept of octal numbers in computer systems.

• Computer Memory
• Binary Operations
• Decimal to Binary
• Octal Representation
• Programming

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1. Module 10 Module 10 Operations on Bits Operations on Bits

2. Whats a bit/byte? Computer memory is made up of hardware that can represent a 0 or a 1. Each individual 0 or 1 is a bit (Binary digIT) C provides operators for manipulating these individual bits Bits are commonly group in sets of 8 known as a byte Bits in a byte are numbered from right-to-left starting with bit 0 Bit 0 is known as the least significant bit (lsb) www.umbctraining.com @UMBC Training Centers 2012 2

3. Bits and Integers Each bit in a byte represents a power of 2 Bit 0 represents 20 = 1 Bit 1 represents 21 = 2 etc Example: What s the value of 001001102 27 26 25 24 23 22 21 20 128 64 32 16 8 4 2 1 0 0 1 0 0 1 1 0 001001102 = 25 + 22 + 21 = 32 + 4 + 2 = 3810 (in binary!!) www.umbctraining.com @UMBC Training Centers 2012 3

4. Binary Exercises 0 0 1 0 1 1 1 0 0 0 1 0 1 1 0 0 0 0 1 1 1 1 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 1 1 1 0 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 1 1 1 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 1 1 1 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 1 0 = = = = = = = = = = = = = = = = = = ex. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 www.umbctraining.com @UMBC Training Centers 2012 4

5. Binary Exercises Round 2 = = = = = = = = = = = 117 86 158 201 136 250 179 23 100 200 231 Ex. 1 2 3 4 5 6 7 8 9 10 www.umbctraining.com @UMBC Training Centers 2012 5

6. Another way of Solving Another way of converting a Decimal to Binary 68 2 = 34 R 0 34 2 = 17 R 0 17 2 = 8 R 1 8 2 = 4 R 0 4 2 = 2 R 0 2 2 = 1 R 0 1 2 = 0 R 1 1 0 0 0 1 0 0 = 68 (base 10) Routinely done in programming www.umbctraining.com @UMBC Training Centers 2012 6

7. Octal Octal = base 8 A group of 3 bits is one octal digit Octal constants start with 0 Binary Octal Decimal 000 0 0 001 1 1 010 2 2 011 3 3 100 4 4 101 5 5 110 6 6 111 7 7 www.umbctraining.com @UMBC Training Centers 2012 7

8. Octal to Decimal www.umbctraining.com @UMBC Training Centers 2012 8

9. Another way of Solving www.umbctraining.com @UMBC Training Centers 2012 9

10. Octal in Applications www.umbctraining.com @UMBC Training Centers 2012 10

11. Exercises www.umbctraining.com @UMBC Training Centers 2012 11

12. Hexadecimal Hexadecimal = base 16 A group of 4 bits is one hex digit Hex constants start with 0x Binary Hex Decimal Binary Hex Decimal 0000 0 0 1000 8 8 0001 1 1 1001 9 9 0010 2 2 1010 A 10 0011 3 3 1011 B 11 0100 4 4 1100 C 12 0101 5 5 1101 D 13 0110 6 6 1110 E 14 0111 7 7 1111 F 15 www.umbctraining.com @UMBC Training Centers 2012 12

13. Hexadecimal to Decimal www.umbctraining.com @UMBC Training Centers 2012 13

14. Hexadecimal in an Application www.umbctraining.com @UMBC Training Centers 2012 14

15. Another way of Solving www.umbctraining.com @UMBC Training Centers 2012 15

16. Exercises www.umbctraining.com @UMBC Training Centers 2012 16

17. Why Binary , Octal and Hex? Translate the following 32-bit integer 00101010011101101000110110100101 Octal 000 101 010 011 101 101 000 110 110 100 101 0 5 2 3 5 5 0 6 6 4 5 Hexadecimal 0010 1010 0111 0110 1000 1101 1010 0101 2 A 7 6 8 D A 5 www.umbctraining.com @UMBC Training Centers 2012 17

18. Twos Complement To store both positive and negative values, the left-most bit is the designated sign bit Sign bit = 0 indicates a positive value Sign bit = 1 indicates a negative value To store a negative number, store its absolute value, add 1, then complement the result. To store 5, first store 5 = 00000101 Then complement = 11111010 Then add 1, = 11111011 www.umbctraining.com @UMBC Training Centers 2012 18

19. Twos Compliment Visually www.umbctraining.com @UMBC Training Centers 2012 19

20. Solving for Twos Compliment www.umbctraining.com @UMBC Training Centers 2012 20

21. More twos complement Largest value in a byte: 01111111 = 127 Smallest value in a byte: 10000000 = -128 -1 in a byte = 11111111 Smallest value in a 4-byte (32-bit) integer 10000000000000000000000000000000 -232 1 = -2,147,483,648 Largest value in 4-byte (32 bit) integer 01111111111111111111111111111111 231 1 = 2, 147, 483, 647 www.umbctraining.com @UMBC Training Centers 2012 21

22. Exercises Integer Representation Exercises Submit it to your submission folder www.umbctraining.com @UMBC Training Centers 2012 22

23. Bitwise Operators Apply to any integer type or to characters Result is 0 or 1 Apply to an integer bit-by-bit www.umbctraining.com @UMBC Training Centers 2012 23

24. Bitwise AND & (AND) short w1 = 25, w2 = 77; short w3 = w1 & w2; A B A & B 0 0 0 w1 = 0000000000011001 w2 = 0000000001001101 ---------------------- w3 = 0000000000001001 = 9 0 1 0 1 0 0 1 1 1 Notes: a & 1 = a ANDing a bit with 1 retains the value of the bit a & 0 = 0 ANDing a bit with 0 sets the bit to 0 www.umbctraining.com @UMBC Training Centers 2012 24

25. Masking Using the AND operator to select bits. char c = 45; c = c & 0x7; //0x means Octal c = 01001101 0x7 = 00000111 c & 0x7 = 00000101 www.umbctraining.com @UMBC Training Centers 2012 25

26. Lets Take a Look C::B Program 12-1 GUESS the output FIRST Then go ahead and run to confirm your answer www.umbctraining.com @UMBC Training Centers 2012 26

27. Bitwise OR | (OR) short w1 = 0431, w2 = 0152; short w3 = w1 | w2; A B A | B 0 0 0 w1 = 100 011 001 w2 = 001 101 010 ----------------- w3 = 101 111 011 = 0573 0 1 1 1 0 1 1 1 1 Notes: a | 1 = 1 ORing a bit with 1 sets the bit to 1 a | 0 = a ORing a bit with 0 retains the value of the bit www.umbctraining.com @UMBC Training Centers 2012 27

28. Turning Bits On Bits may be turned on (set to 1) by ORing short w1 = 0431; w1 = w1 | 07; w1 = 100 011 001 07 = 000 000 111 w1 | 7 = 100 011 111 www.umbctraining.com @UMBC Training Centers 2012 28

29. Bitwise XOR ^ (XOR) short w1 = 0536, w2 = 0266; short w3 = w1 ^ w2; A B A ^ B 0 0 0 w1 = 101 011 110 w2 = 010 110 110 ----------------- w3 = 111 101 000 = 0750 0 1 1 1 0 1 1 1 0 Notes: 1 ^ 1 = 0 and 0 ^ 1 = 1 XORing a bit with 1 changes the bit 0 ^ 0 = 0 and 1 ^ 0 = 1 XORing a bit with 0 retains the value of the bit www.umbctraining.com www.umbctraining.com @UMBC Training Centers 2012 UMBC Training Centers, LLC 29 29

30. Bitwise Complement ~ (1 s complement) short w1 = 0122457; short w2 = ~w1; A ~A 0 1 w1 = 001 010 010 100 101 111 = 0122457 w2 = 110 101 101 011 010 000 = 0055320 1 0 Notes: ~0 = -1 regardless of the number of bits in an int www.umbctraining.com @UMBC Training Centers 2012 30

31. Lets Take a Look C::B- Program12-2 Predict output first, run it to confirm Note: warnings from line 15 suggest parens Change to use parens around 1st two operands different results! Note precedence of bitwise operators Pg 440 -> Kochan book Redo parens around last 2 operand to get original results3 www.umbctraining.com @UMBC Training Centers 2012 31

32. Exercise PP2-14.docx Complete Submit in your submit folder www.umbctraining.com @UMBC Training Centers 2012 32

33. Left-Shift Operator The << (left shift) operator literally moves bits in a value to the left. The left-most (high- order) bits are lost and 0 s replace the right most (low-order) bits which are shifted. char c = 0x43; char d = c << 2; c = 0100 0011 d = 0000 1100 www.umbctraining.com @UMBC Training Centers 2012 33

34. Notes on Shifting Left-shift is used to multiply by 2N Shifting with a count that is more than 32 bits (number of bits in an int) is undefined www.umbctraining.com @UMBC Training Centers 2012 34

35. Right-shift Operator The >> (right shift) operator literally moves bits in a value to the right. The right-most (low-order) bits are lost. If the value is signed (the default) then copies of the sign-bit fill in the high-order bits. If the value is unsigned , then 0s fill in the high-order bits. www.umbctraining.com @UMBC Training Centers 2012 35

36. Signed Right-Shift signed short w1 = 0x9013, w2 = 0x0322; signed short w3 = w1 >> 3; signed short w4 = w2 >> 2; w1 = 1001 0000 0001 0011 w3 = 1111 0010 0000 0010 w2 = 0000 0011 0010 0010 w4 = 0000 0000 1100 0100 www.umbctraining.com @UMBC Training Centers 2012 36

37. Notes on Shifting Right-shift is used to divide by 2N Shifting with a negative count is undefined www.umbctraining.com @UMBC Training Centers 2012 37

38. Unsigned Right-Shift unsigned short w1 = 0x9013, w2 = 0x0322; unsigned short w3 = w1 >> 3; unsigned short w4 = w2 >> 2; w1 = 1001 0000 0001 0011 w3 = 0001 0010 0000 0010 w2 = 0000 0011 0010 0010 w4 = 0000 0000 1100 0100 www.umbctraining.com @UMBC Training Centers 2012 38

39. Lets take a look C::B Program 12-3 www.umbctraining.com @UMBC Training Centers 2012 39

40. Rotating Bits No rotate operator Need to write a function Similar to shifting Left-Rotate High-order bits that were lost when shifting become the low-order bits Right-Rotate Low-order bits that were lost when shifting become the high-order bits www.umbctraining.com @UMBC Training Centers 2012 40

41. Rotate Examples unsigned short w1 = 0x4356; unsigned short w2 = rightRotate(w1, 3); Unsigned short w3 = 0x9234; unsigned short w4 = leftRotate(w1, 2); w1 = 0100 0011 0101 0110 w2 = 1100 1000 0110 1010 w3 = 1001 0010 0011 0100 w4 = 0100 1000 1101 0010 www.umbctraining.com @UMBC Training Centers 2012 41

42. Lets Take a Look C::B Program 12-4 Notice that shift operators have higher precedence than bit-wise ops Would prefer separate functions www.umbctraining.com @UMBC Training Centers 2012 42

43. Bit Fields Multiple data values can be stored in an int, short Values are set and extracted via mask & shift 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 day 1 - 31 year 0 -99 month 1 - 12 www.umbctraining.com @UMBC Training Centers 2012 43

44. March 11, 2033 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 day 1 - 31 year 0 -99 month 1 - 12 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 day = 11 year = 33 month = 3 www.umbctraining.com @UMBC Training Centers 2012 44

45. Extracting Day from the Byte 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 int ymd = 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 day = 11 year = 33 month = 3 0x1F = 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 ymd & 0x1F = 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 day = ymd & 0x1F; www.umbctraining.com @UMBC Training Centers 2012 45

46. Extracting Month 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 int ymd = 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 day = 11 year = 33 month = 3 0x1E0 = 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 ymd & 0x1E0 = 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 ymd & 0x1E0 >> 5 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 month = ymd & 0x1E0 >> 5; www.umbctraining.com @UMBC Training Centers 2012 46

47. Extracting Year 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 ymd = day = 11 year = 33 month = 3 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0x7E00 = ymd & 0x7E00 = 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 ymd & 0x7E00 >> 9 = 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 year = ymd & 0x7E00 >> 9; www.umbctraining.com @UMBC Training Centers 2012 47

48. Packing Bits 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 // March 11, 2033 short day = 11; short month = 3; short year = 33; 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 0 1 1 day = 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 month = 3 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 year = 33 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 1 1 day = 11 year = 33 month=3 shortymd = (year << 9) | (month << 5) | day; www.umbctraining.com @UMBC Training Centers 2012 48

49. Struct Bit Fields struct ymd { unsigned int year:6; unsigned int month:4; unsigned int day: 5; }; struct ymd birthday; // in main() Access fields using dot notation Fields automatically converted to integers birthday.year = 44; printf( %d\n , birthday.year); www.umbctraining.com @UMBC Training Centers 2012 49

50. Exercises Text (page 297) #1 (12.1, 12.2, 12.3 in HEX, 12.4 in octal) # Text #5 + 7 Ex12-5_7.docx bitTest and bitSet 2 lines of code bigGet - extract bits from an unsigned int. e.g. extractBits( x, 3, 5) extracts bits 3-7 bit is lsb Ex12-6.docx bitPatternMatch( x, pattern, n ) are the rightmost N bits of pattern in x Change bit orientation so that bit 0 is lsb Assume 32-bit integers Ex1-Bytes.docx -Extract byte N Challenge - replace value in byte N Ex2-Fields.docx breaking apart packed data www.umbctraining.com @UMBC Training Centers 2012 50

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