1. NUMBER REPRESENTATION CHAPTER 3 – part 3

2. ONE’S COMPLEMENT REPRESENTATION CHAPTER 3 – part 3

3. Integer Representation UnsignedSigned Sign and Magnitude One’s Complement Two’s Complement

4. One’s complement integers

5. Table 3.5 Range of One’s complement integers # of Bits # of Bits --------- 8 16 32 Range-------------------------------------------------------  127  0  32767  0   0 +0 +127 +0 +32767 +0 +2,147,483,647

6. One’s complement integers There are two 0s in There are two 0s in One’s complement representation: positive and negative. In an 8-bit allocation: +0  00000000 -0  10000000 There are two 0s in There are two 0s in One’s complement representation: positive and negative. In an 8-bit allocation: +0  00000000 -0  10000000

7. One’s complement integers In one’’s complement representation, the leftmost bit defines the sign of the number. If it is 0, the number is positive. If it is 0, the number is positive. If it is 1, the number is negative In one’’s complement representation, the leftmost bit defines the sign of the number. If it is 0, the number is positive. If it is 0, the number is positive. If it is 1, the number is negative

8. One’s complement integers Storing One’s complement integer process: 1. The integer is changed to binary, (the sign is ignored). 2. 0s are added to the left of the number to make a total of N 3. bits 4. If the sign is positive, no more action is needed. If the sign is negative, every bit is complemented

9. Store +7 in an 8-bit memory location using one’s complement representation.  The integer is changed to binary (111).  Add 5 0s to make a total of N (8) bits, 00000111.  The sign is positive so no more action is needed Example 4 Solution

10. Example 5 Solution

11. One’s complement integers Table 3.6 Example of storing one’s complement integers in two computers Decimal Decimal ------------ +7 −7 +124 −124 +24,760 −24,760 8-bit allocation ------------ 00000111 11111000 01111100 10000011 overflow 16-bit allocation ------------------------------ 0000000000000111 1111111111111000 0000000001111100 1111111110000011 0110000010111000 1001111101000111

12. One’s complement integers

13. Example 6 Solution

14. The one’s complement representation is not used now by computers because:  Operations: such as subtraction and addition is not straightforward for this representation.  Uncomfortable in programming: because there are two 0s in this representation One’s complement Applications

15. However.. The advantage of this representation is: 1. It’s the foundation of the next representation(Two’s complement) 2. It has properties that make it interesting for data communication applications such as error detection and correction One’s complement Applications

16. TWO’S COMPLEMENT REPRESENTATION

17. Integer Representation UnsignedSigned Sign and Magnitude One’s Complement Two’s Complement

18. Two’s complement integers It’s… The most common The most important The most widely used representation today

19. Two’s complement integers Table 3.7 Range of Two’s complement integers # of Bits # of Bits --------- 8 16 32 Range-------------------------------------------------------  128  0  32768  0   0 +0 +127 +0 +32767 +0 +2,147,483,647

20. Two’s complement integers There is Only one 0 in There is Only one 0 in Two’s complementrepresentation In an 8-bit allocation: 0  00000000 There is Only one 0 in There is Only one 0 in Two’s complementrepresentation In an 8-bit allocation: 0  00000000

21. Two’s complement integers In two’s complement representation, the leftmost bit defines the sign of the number. If it is 0, the number is positive. If it is 0, the number is positive. If it is 1, the number is negative In two’s complement representation, the leftmost bit defines the sign of the number. If it is 0, the number is positive. If it is 0, the number is positive. If it is 1, the number is negative

22. Two’s complement integers Storing One’s complement integer process: 1. The number is changed to binary, (the sign is ignored). 2. 0s are added to the left of the number to make a total of N bits 3. If the sign is positive, no more action is needed. If the sign is negative, leave all the rightmost 0’s and the first 1 unchanged. Then complement the rest of the bits.

23. Store +7 in an 8-bit memory location using two’s complement representation.  The integer is changed to binary (111).  Add 5 0s to make a total of N (8) bits, 00000111.  The sign is positive so no more action is needed Example 4 Solution

24. Example 5 Solution

25. Two’s complement integers Table 3.8 Example of storing two’s complement integers in two computers Decimal Decimal ------------ +7 −7 +124 −124 +24,760 −24,760 8-bit allocation ------------ 00000111 11111001 01111100 10000100 overflow 16-bit allocation ------------------------------ 0000000000000111 1111111111111001 0000000001111100 1111111110000100 0110000010111000 1001111101001000

26. Two’s complement integers

27. Example 6 Solution

28. The two’s complement representation is the standard representation used for storing integers by computers today because it makes the operations simple. Two’s complement Applications

29. SUMMARY OF INTEGER REPRESENTATION Unsigned Sign & magnitude One’s complement Two’s complement

30. One’s complement Sign & magnitude Unsigned Contents of Memory +0 00000 +1 10001 +2 20010 +3 30011 +4 40100 +5 50101 +6 60110 +7 70111 -8-7-081000 -7-691001 -6-5-2101010 -5-4-3111011 -4-3-4121100 -3-2-5131101 -2-6141110 -0-7151111

31. EXCESS SYSTEM

32. Excess System

33. Example

34. Excess System Representation To represent a number in Excess, use the following procedure:  ‰ Add the magic number to the integer  ‰ Change the result to binary and add 0s so that there is a total of N bits

35. Excess System Representation

36. Excess System Interpretation To interpret a number in Excess, use the following procedure:  ‰Change the number to decimal  Subtract the magic number from the integer

37. Excess System Interpretation

38. FLOATING POINT REPRESENTATION

39. Floating point representation Floating point number = a number containing an integer & a fraction Example: 14.234 IntegerFraction

40. Convert from Floating point number to binary 1. Convert the integer part to binary 2. ‰Convert the fraction to binary 3. ‰Put a decimal point between the two parts

41. Convert from Floating point number to binary Example Transform the fraction 0.875 to binary Answer: Write the fraction at the left corner. Multiply the number continuously by 2 and extract the integer part as the binary digit. Stop when the number is 0.0. fraction binary 0.875  1.750  1.50  1.0  0.0 0. 1 1 1

42. Convert from Floating point number to binary Example Transform the fraction 0.4 to a binary of 6 bits. Answer: Write the fraction at the left cornet. Multiply the number continuously by 2 and extract the integer part as the binary digit. You can never get the exact binary representation. Stop when you have 6 bits fraction binary 0.4  0.8  1.6  1.2  0.4  0.8  1.6 0. 0 1 1 0 0 1

43. NORMALIZATION

44. Normalization A fraction is normalized so that operations are simpler Normalization: the moving of the decimal point so that there is only one 1 to the left of the decimal point.

45. Normalization NormaliazedMoveOriginal Number + 2 6 x 1.01000111001 6+ 1010001.11001 -2 2 x 1.110011 2- 111.000011 +2 -6 x 1.110016+ 0.00000111001 -2 -3 x 1.1100113- 0.001110011

46. Normalization After the number is normalized we store 3 pieces of information about it: SIGN EXPONENT MANTISSA Example : + 2 6 x 1.0001110101 SIGNEXPONENTMANTISSA

47. IEEE standard for floating point representation

48. Single-Precesion Representation The procedure of storing a normalized floating point number using single precession format is as follow: 1. Store the sign as 0 (positive) or 1 (negative) 2. Store the exponent (power of 2) as Excess_127 3. Store the mantissa as unsigned integer

49. EXAMPLE  Show the representation of the normalized number + 2 6 x 1.01000111001 ANSWER: The sign is positive. The Excess_127 representation of the exponent is 133. You add extra 0s on the right to make it 23 bits. The number in memory is stored as: 0 10000101 01000111001000000000000 SIGNEXPONENTMANTISSA

50. Example of floating point representation MantissaExponentSignNumber 11000011000000000000000100000011-2 2 x 1.11000011 11001000000000000000000011110010+2 -6 x 1.11001 11001100000000000000000011111001-2 -3 x 1.110011

51. Floating point Interpretation for single precision the following procedure interprets a 32-bit floating-point number stored in memory 1. Use the leftmost bit as the sign 2. Change the next 8 bit to decimal and subtract 127 from it. (this is the exponent) 3. Add 1 and a decimal point to the next 23 bits (ignore any extra 0’s in the right) 4. Move the decimal point to the correct position using the value of the exponent 5. Change the whole part to decimal 6. Change the fraction part to decimal 7. Combine the whole and the fraction parts

52. Example Interpret the following 32-bit floating-point number 1 01111100 11001100000000000000000 ANSWER : The sign is negative. The exponent is –3 (124 –127). The number after normalization is - 2 -3 x 1.110011