1. Dr. ClincyLecture 2 Slide 1 CS 3501 - Chapter 2 (1 of 5) Dr. Clincy Professor of CS Note: Do not study chapter 2’s appendix (the topics will be covered in the communications portion of this course)

2. 2 Chapter 2 Objectives The architecture of a computer depends on how it represents numbers, characters and control info. Understand the fundamental concepts of number systems. Understand the fundamental concepts of floating-point representation. Gain familiarity with the most popular character codes. Understand the concepts of error detecting and correcting.

3. 3 Introduction A bit is the most basic unit of information in a computer. –Sometimes these states are “high” or “low” voltage or “on” or “off..” A byte is a group of eight bits. –A byte is the smallest possible addressable unit of computer storage. A word is a contiguous group of bytes. –Words can be any number of bits or bytes. –Word sizes of 16, 32, or 64 bits are most common. –In a word-addressable system, a word is the smallest addressable unit of storage. A group of four bits is called a nibble. –Bytes, therefore, consist of two nibbles: a “high-order nibble,” and a “low-order” nibble.

4. Dr. ClincyLecture4 Positional Numbering Systems Review – Base 10 Numbers (Decimal) Base-10 The decimal number system is based on power of the base 10. For example, for the number 1259, the 9 is in the 10^0 column - 1s column the 5 is in the 10^1 column - 10s column the 2 is in the 10^2 column - 100s column the 1 is in the 10^3 column - 1000s column 1259 is 9 X 1 = 9 + 5 X 10 = 50 + 2 X 100 = 200 + 1 X 1000 = 1000 ----- 1259

5. Dr. ClincyLecture5 Positional Numbering Systems Introducing Base 2 (Binary) and Base 16 (Hex) Number Systems Base-2 (Binary) The Binary number system uses the same mechanism and concept however, the base is 2 versus 10 The place values for binary are based on powers of the base 2: … 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 128 64 32 16 8 4 2 1 Base-16 (Hex) The hexadecimal number system is based 16, and uses the same mechanisms and conversion routines we have already examined. The place values for hexadecimal are based on powers of the base 16 The digits for 10-15 are the letters A - F (A is 10, …….., F is 15) …….. 16^3 16^2 16^1 16^0 4096 256 16 1

6. Dr. ClincyLecture6 5-bit Binary Number System 2 4, 2 3, 2 2, 2 1, 2 0 16, 8, 4, 2, 1

7. Dr. ClincyLecture7 Different Number Systems Base-10 (Decimal) – what are the characters ? Example = 659 Base-2 (Binary) – what are the characters ? Example = 1101 Base-16 (Hex) – what are the characters ? Example = AE Base-8 (Octal) – what are the characters ? Example = 73

8. 8 Converting 190 to base 3... –3 5 = 243 is too large, so we try 3 4 = 81. And 2 times 81 doesn’t exceed 190 –The last power of 3, 3 0 = 1, is our last choice, and it gives us a difference of zero. –Our result, reading from top to bottom is: 190 10 = 21001 3 Converting Between Bases – Subtraction Method

9. 9 Converting 190 to base 3... –Continue in this way until the quotient is zero. –In the final calculation, we note that 3 divides 2 zero times with a remainder of 2. –Our result, reading from bottom to top is: 190 10 = 21001 3 Converting Between Bases –Division Method

10. Dr. ClincyLecture10 Converting from Binary to Decimal So, the binary number 10110011 can be converted to a decimal number 1 X 1 = 1 (right most bit or position) 1 X 2 = 2 0 X 4 = 0 0 X 8 = 0 1 X 16 = 16 1 X 32 = 32 0 X 64 = 0 1 X 128 = 128 (left most bit or position) ------ 179 in decimal

11. Dr. ClincyLecture11 Converting from Decimal to Binary To convert from decimal to some other number system requires a different method called the division/remainder method. The idea is to repeatedly divide the decimal number and resulting quotients by the number system’s base. The answer will be the remainders. Example: convert 155 to binary (Start from the top and work down) 155/2 Q = 77, R = 1 (Start) 77/2 Q = 38, R = 1 38/2 Q = 19, R = 0 19/2 Q = 9, R = 1 9/2 Q = 4, R = 1 4/2 Q = 2, R = 0 2/2 Q = 1, R = 0 1/2 Q = 0, R = 1 (Stop) Answer is 10011011. Be careful to place the digits in the correct order.

12. 12 Why Decimal, Binary and Hex ? Give subscripts for Decimal, Binary, Hex, Octal Converting Between Bases

13. 13 Using groups of hextets, the binary number 11010100011011 2 (= 13595 10 ) in hexadecimal is: Octal (base 8) values are derived from binary by using groups of three bits (8 = 2 3 ): Octal was very useful when computers used six-bit words. If the number of bits is not a multiple of 4, pad on the left with zeros. Converting Between Bases of Power 2

14. 14 Fractional decimal values have nonzero digits to the right of the decimal point. Fractional values of other radix systems have nonzero digits to the right of the radix point. Numerals to the right of a radix point represent negative powers of the radix: 0.47 10 = 4  10 -1 + 7  10 -2 0.11 2 = 1  2 -1 + 1  2 -2 = ½ + ¼ = 0.5 + 0.25 = 0.75 Converting Between Bases

15. 15 The calculation to the right is an example of using the subtraction method to convert the decimal 0.8125 to binary. –Our result, reading from top to bottom is: 0.8125 10 = 0.1101 2 –Of course, this method works with any base, not just binary. Subtraction - Converting Between Bases

16. 16 Converting 0.8125 to binary... –Multiplication Method: You are finished when the product is zero, or until you have reached the desired number of binary places. –Our result, reading from top to bottom is: 0.8125 10 = 0.1101 2 –This method also works with any base. Just use the target radix as the multiplier. Multiplication - Converting Between Bases

17. Dr. ClincyLecture17 Converting Number Systems

18. Dr. ClincyLecture 3 Slide 18 CS 3510 - Chapter 2 (2 of 5) Dr. Clincy Professor of CS

19. Dr. ClincyLecture19 Addition Dr. Clincy19

20. Dr. ClincyLecture20 Addition & Subtraction Dr. Clincy20

21. What about multiplication in base 2 By hand - For unsigned case, very similar to base-10 multiplication Dr. ClincyLecture21Dr. Clincy21

22. Dr. ClincyLecture22 Division Dr. Clincy22

23. Dr. ClincyLecture23 Computers can not SUBTRACT Therefore, need approaches to represent “signed” numbers

24. Dr. ClincyLecture24 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1+ 1- 2+ 3+ 4+ 5+ 6+ 7+ 2- 3- 4- 5- 6- 7- 8- 0+ 0- 1+ 2+ 3+ 4+ 5+ 6+ 7+ 0+ 7- 6- 5- 4- 3- 2- 1- 0- 1+ 2+ 3+ 4+ 5+ 6+ 7+ 0+ 7- 6- 5- 4- 3- 2- 1- b 3 b 2 b 1 b 0 Sign and magnitude 1's complement2's complement BValues represented. Changing only b 3 makes +/-Complementing makes +/- Complementing plus 1 makes +/- More used and more efficient NOTE: S-M can be used to “represent” signed numbers however, 1’s and 2’s complement can also be used for subtraction

25. Dr. ClincyLecture25 5-bit Binary Number System with SIGN X 4, X 3, X 2, X 1, X 0 Sign

26. 26 Example: –Using signed magnitude binary arithmetic, find the sum of 75 and 46. Once we have worked our way through all eight bits, we are done. In this example, we were careful to pick two values whose sum would fit into seven bits. If that is not the case, we have a problem. Sign Magnitude Approach Dr. ClincyLecture

27. 27 Example: –Using signed magnitude binary arithmetic, find the sum of 107 and 46. We see that the carry from the seventh bit overflows and is discarded, giving us the erroneous result: 107 + 46 = 25. Sign Magnitude Approach Dr. ClincyLecture

28. 28 The signs in signed magnitude representation work just like the signs in pencil and paper arithmetic. –Example: Using signed magnitude binary arithmetic, find the sum of - 46 and - 25. Because the signs are the same, all we do is add the numbers and supply the negative sign when we are done. Sign Magnitude Approach Dr. ClincyLecture

29. 29 Mixed sign addition (or subtraction) is done the same way; pencil and paper arithmetic –Example: Using signed magnitude binary arithmetic, find the sum of 46 and - 25. The sign of the result gets the sign of the number that is larger. –Note the “borrows” from the second and sixth bits. Sign Magnitude Approach Dr. ClincyLecture

30. 30 Signed magnitude representation is easy for people to understand, but it requires complicated computer hardware. Another disadvantage of signed magnitude is that it allows two different representations for zero: positive zero and negative zero. For these reasons (among others) computers systems employ 1’s and 2’s complement approaches Sign Magnitude Approach Dr. ClincyLecture 9 and 10

31. 31 For example, using 8-bit one’s complement representation: + 3 is : 00000011 - 3 is: 11111100 In one’s complement representation, as with signed magnitude, negative values are indicated by a 1 in the high order bit. Complement systems are useful because they eliminate the need for subtraction. One’s Complement Approach Dr. ClincyLecture 9 and 10

32. 32 With one’s complement addition, the carry bit is “carried around” and added to the sum. –Example: Using one’s complement binary arithmetic, find the sum of 48 and -19 We note that 19 in binary is : 00010011, So -19 in one’s complement is: 11101100. One’s Complement Approach Dr. ClincyLecture 9 and 10

33. Dr. ClincyLecture 9 and 1033 One’s Complement - Binary Addition 1010 (neg 5) +0010 (pos 2) 1100 (neg 3) 1101 (neg 2) +0111 (pos 7) 10100 (overflow – add the 1 back) 0101 (pos 5) Recall complement 0011

34. Dr. ClincyLecture 9 and 1034 1’s Complement

35. Dr. ClincyLecture 9 and 1035 1’s Complement

36. 36 Although the “end carry around” adds some complexity, one’s complement is simpler to implement than signed magnitude. But it still has the disadvantage of having two different representations for zero: positive zero and negative zero. Two’s complement solves this problem. One’s Complement Approach Dr. ClincyLecture 9 and 10

37. 37 To express a value in two’s complement representation: –If the number is positive, just convert it to binary and you’re done. –If the number is negative, find the one’s complement of the number and then add 1. Example: –In 8-bit binary, 3 is: 00000011 –-3 using one’s complement representation is: 11111100 –Adding 1 gives us -3 in two’s complement form: 11111101. Two’s Complement Approach Dr. ClincyLecture 9 and 10

38. 38 With two’s complement arithmetic, all we do is add our two binary numbers. Just discard any carries emitting from the high order bit. We note that 19 in binary is: 00010011, So -19 using one’s complement is: 11101100, So -19 using two’s complement is: 11101101. –Example: Using two’s complement binary arithmetic, find the sum of 48 and - 19. Two’s Complement Approach Dr. ClincyLecture 9 and 10

39. 39 Example: –Using two’s complement binary arithmetic, find the sum of 107 and 46. We see that the nonzero carry from the seventh bit overflows into the sign bit, giving us the erroneous result: 107 + 46 = -103. Sign-bit: Carry-in of 1 and Carry-out of 0 But overflow into the sign bit does not always mean that we have an error. Two’s Complement Approach Rule for detecting signed two’s complement overflow: When the “carry in” and the “carry out” of the sign bit differ, overflow has occurred. If the carry into the sign bit equals the carry out of the sign bit, no overflow has occurred. Dr. ClincyLecture 9 and 10

40. 40 Example: –Using two’s complement binary arithmetic, find the sum of 23 and -9. –We see that there is carry into the sign bit and carry out. The final result is correct: 23 + (-9) = 14. –Sign-bit: Carry-in of 1 and Carry-out of 0 Rule for detecting signed two’s complement overflow: When the “carry in” and the “carry out” of the sign bit differ, overflow has occurred. If the carry into the sign bit equals the carry out of the sign bit, no overflow has occurred. Two’s Complement Approach Dr. ClincyLecture 9 and 10