1. Dale Roberts Department of Computer and Information Science, School of Science, IUPUI CSCI 230 Information Representation: Negative and Floating Point Representation Dale Roberts, Lecturer IUPUIdroberts@cs.iupui.edu

2. Dale Roberts Negative Numbers in Binary Four different representation schemes are used for negative numbers 1. Signed Magnitude Left most bit (LMB) is the sign bit : 0  positive (+) 1  negative (-) Remaining bits hold absolute magnitude Example: 2 10  0000 0010 b -2 10  1000 0010 b Q: 0000 0000 = ? 1000 0000 = ? Try, 1000 0100 b =-4 10

3. Dale Roberts 2.One’s Complement – –Left most bit is the sign bit : 0  positive (+) 1  negative (-) – –The magnitude is Complemented Example: 2 10  0 000 0010 b -2 10  1 111 1101 b Exercise: try - 4 10 using 1’s Complement Q: 0000 0000 = ? 1111 1111 = ? 1111 1111 = ? Solution: 4 10 = 0 000 0100 b -4 10 = 111 1011 b 1

4. Dale Roberts Negative Numbers in Binary (cont.) 3.2’s Complement Sign bit same as above Magnitude is Complemented first and a “1” is added to the Complemented digits Example: 2 10  0 000 0010 b 1’s Complement  1 111 1101 b + 1 -2 10  1 111 1110 b 7 10  1’s Complement  + 1 -7 10  Exercise: try -7 10 using 2’s Complement 0000 0111 b 1111 1000 b 1111 1001 b

5. Dale Roberts Negative Numbers in Binary (cont.) 7 10 = 0000 0111 b 3 10 = 0000 0011 b 1’s complement 1111 1100 b 2’s complement 1111 1101 b  -3 10 7+(-3)  0000 0111 +1111 1101 +1111 1101 Example: 7+(-3) [hint]: A – B = A + (~B) +1 1 1111 111 carry ignore 1 0000 0100  0000 0100  4 10

6. Dale Roberts Three Representation of Signed Integer

7. Dale Roberts Negative Numbers in Binary (cont.) 4. Excess Representation – – For a given fixed number of bits the range is remapped such that roughly half the numbers are negative and half are positive. Example: (as left) Excess – 8 notation for 4 bit numbers Binary value = 8 + excess-8 value MSB can be used as a sign bit, but If MSB =1, positive number If MSB =0, negative number Numbers Binary Value Notation Excess – 8 Value 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 -8 -7 -6 -5 -4 -3 -2 0 1 2 3 4 5 6 7

8. Dale Roberts Fundamental Data Type 2 byte unsigned (Default type is int ) 2 byte int 0000 0000 0000 0000 (  0 D ) 0000 0000 0000 0001 (  1 D ) 0000 0000 0000 0010 (  2 D ) …. 0111 1111 1111 1111 (  32767 D  2 15 -1) 1000 0000 0000 0000 (  32768 D  2 15 ) …. 1111 1111 1111 1111 (  2 16 –1) 1000 0000 0000 0000 (  -32768 D  - 2 15 ) 1000 0000 0000 0001 (  -32767 D  - 2 15 +1) …. 1111 1111 1111 1110 (  - 2 D ) 1111 1111 1111 1111 (  - 1 D ) 0000 0000 0000 0000 (  0 D ) 0000 0000 0000 0001 (  1 D ) 0000 0000 0000 0010 (  2 D ) …. 0111 1111 1111 1111 (  32767 D  2 15 -1) With vs. without using sign bit With vs. without using sign bit For a 16 bit binary pattern:

9. Dale Roberts Fundamental Data Type Four Data Types in C Four Data Types in C (assume 2’s complement, byte machine) Data Type Abbreviation Size (byte) Range char 1-128 ~ 127 unsigned char10 ~ 255 int 2 or 4-2 15 ~ 2 15 -1 or -2 31 ~ 2 31 -1 unsigned int unsigned 2 or 40 ~ 65535 or 0 ~ 2 32 -1 short int short 2-32768 ~ 32767 unsigned short int unsigned short 20 ~ 65535 long int long 4-2 31 ~ 2 31 -1 unsigned long int unsigned long 40 ~ 2 32 -1 float4 double8 Note:2 7 = 128, 2 15 =32768, 2 15 = 2147483648 Complex and double complex are not available

10. Dale Roberts Fractional Numbers Examples: 456.78 10 = 4 x 10 2 + 5 x 10 1 + 6 x 10 0 + 7 x 10 -1 +8 x 10 -2 1011.11 2 = 1 x 2 3 + 0 x 2 2 + 1 x 2 1 + 1 x 2 0 + 1 x 2 -1 + 1 x 2 -2 = 8 + 0 + 2 + 1 + 1/2 + ¼ = 11 + 0.5 + 0.25 = 11.75 10 Conversion from binary number system to decimal system Examples: 111.11 2 = 1 x 2 2 + 1 x 2 1 + 1 x 2 0 + 1 x 2 -1 + 1 x 2 -2 = 4 + 2 + 1 + 1/2 + ¼ = 7.75 10 Examples: 11.011 2 2 2 2 1 2 0 2 -1 2 -2 2 - 3 4 2 1 ½ ¼ 1/8 2 1 0 -1 -2 -3 xxxx

11. Dale Roberts Conversion from decimal number system to binary system Examples: 7.75 10 = (?) 2 1. Conversion of the integer part: same as before – repeated division by 2 7 / 2 = 3 (Q), 1 (R)  3 / 2 = 1 (Q), 1 (R)  1 / 2 = 0 (Q), 1 (R) 7 10 = 111 2 2. Conversion of the fractional part: perform a repeated multiplication by 2 and extract the integer part of the result 0.75 x 2 =1.50  extract 1 0.5 x 2 = 1.0  extract 1 0.75 10 = 0.11 2 0.0  stop  Combine the results from integer and fractional part, 7.75 10 = 111.11 2  Combine the results from integer and fractional part, 7.75 10 = 111.11 2 How about choose some of Examples: try 5.625 B write in the same order 421 1/21/41/8 =0.5 =0.25=0.125

12. Dale Roberts Fractional Numbers (cont.) Exercise 1: Convert (0.625) 10 to its binary form Exercise 2: Convert (0.6) 10 to its binary form Solution: Solution: 0.625 x 2 = 1.25  extract 1 0.25 x 2 = 0.5  extract 0 0.5 x 2 = 1.0  extract 1 0.0  stop  (0.625) 10 = (0.101) 2 0.6 x 2 = 1.2  extract 1 0.2 x 2 = 0.4  extract 0 0.4 x 2 = 0.8  extract 0 0.8 x 2 = 1.6  extract 1 0.6 x 2 =   (0.6) 10 = (0.1001 1001 1001 …) 2

13. Dale Roberts Fractional Numbers (cont.) Exercise 3: Convert (0.8125) 10 to its binary form Solution: 0.8125 x 2 = 1.625  extract 1 0.625 x 2 = 1.25  extract 1 0.25 x 2 = 0.5  extract 0 0.5 x 2 = 1.0  extract 1 0.0  stop  (0.8125) 10 = (0.1101) 2

14. Dale Roberts Fractional Numbers (cont.) Errors One source of error in the computations is due to back and forth conversions between decimal and binary formats Example: (0.6) 10 + (0.6) 10 = 1.2 10 Since (0.6) 10 = (0.1001 1001 1001 …) 2 Lets assume a 8-bit representation: (0.6) 10 = (0.1001 1001) 2, therefore 0.60.10011001 + 0.6  +0.10011001 1.00110010 Lets reconvert to decimal system: (1.00110010) b = 1 x 2 0 + 0 x 2 -1 + 0 x 2 -2 + 1 x 2 -3 + 1 x 2 -4 + 0 x 2 -5 + 0 x 2 -6 + 1 x 2 -7 + 0 x 2 -8 = 1 + 1/8 + 1/16 + 1/128 = 1.1953125  Error = 1.2 – 1.1953125 = 0.0046875 = 0.0046875

15. Dale Roberts If x is a real number then its normal form representation is: x = f Base E where f : mantissa E: exponent exponent Example: 125.32 10 = 0.12532 10 3 mantissa - 125.32 10 = - 0.12532 10 3 0.0546 10 = 0.546 10 –1 The mantissa is normalized, so the digit after the fractional point is non-zero. If needed the mantissa should be shifted appropriately to make the first digit (after the fractional point) to be non-zero & the exponent is properly adjusted. Floating Point Number Representation

16. Dale Roberts Example: 134.15 10 = x 10 0.0021 10 = x 10 101.11 B = 0.011 B = AB.CD H = 0.00AC H = 0.13415 0.21 3 -2

17. Dale Roberts Assume we use 16-bit binary pattern for normalized binary form based on the following convention (MSB to LSB) Sign of mantissa ( ± )= left most bit (where 0: +; 1: - ) Mantissa (f)= next 11 bits Sign of exponent ( ± )= next bit (where 0: +; 1: - ) Exponent (E) = next three bits x = ± f Base ± E LSBMSB + : 0 - : 1 E : converted to binary, b 1 b 2 b 3 ?1?1 ?2?2 ?3?3 ?4?4 ? 11 ? 10 ?9?9 ?8?8 ?7?7 ?5?5 ?6?6 f = 0.? 1 ? 2 ? 3 ? 4 …? 11 ? 12 …? 15 b1b1 b2b2 b3b3 + : 0 - : 1

18. Dale Roberts Question: How the computer expresses the 16-bit approximation of 1110.111010111111 in normalized binary form using the following convention Sign of mantissa = left most bit (where 0: +; 1: - ) Mantissa = next 11 bits Sign of exponent = next bit (where 0: +; 1: - ) Exponent = next three bitsAnswer: Step 1: Normalization 1110.111010111111 = + 0.1110111010111111 * 2 +4 Step 2: “Plant” 16 bits the 16 bit floating point representation is 0 11101110101 0 100 Floating Point Number Representation exponent 3 bits sign 1 bit mantissa 11 bits sign 1 bit

19. Dale Roberts Question: Interpret the normalized binary number 0111 0000 0000 1010 B using the convention mentioned Sign of mantissa = left most bit (where 0: +; 1: - ) Mantissa = next 11 bits Sign of exponent = next bit (where 0: +; 1: - ) Exponent = next three bits find its decimal equivalent. Answer: 0 11100000000 1 010 B = 0.111 B * 2 -2 = 7/32 = 0.21875 D

20. Dale Roberts Real Life Example: IEEE 754 IEEE Standard 754 is the representation of floating point used on most computers. Single precision (float) is 32 bits or 4 bytes with the following configuration. 1 sign bit 8 exponent 23 fraction The sign field is 0 for positive or 1 for negative The exponent field has a bias of 127, meaning that 127 is added to the exponent before it’s stored. 2 0 becomes 127, 2 1 becomes 128, 2 -3 becomes 124, etc. In the mantissa, the decimal point is assumed to follow the first ‘1’. Since the first digit is always a ‘1’, a hidden bit is used to representing the bit. The fraction is the 23 bits following the first ‘1’. The fraction really represents a 24 bit mantissa.

21. Dale Roberts Real Life Example: IEEE 754 IEEE 754 Examples: Normalized Numbers 0 1000 0011 0000 0000 0000 0000 0000 000 = 1 x 2 4 = 16 0 0011 0001 0000 0000 0000 0000 0000 000 = 1 x 2 -78 = 3.3087e-24 0 1000 0001 0100 0000 0000 0000 0000 000 = 1.25 x 2 2 = 5

22. Dale Roberts Real Life Example: IEEE 754 Double precision (double) is 64 bites or 8 bytes with the following configuration. 1 sign bit 11 exponent 52 fraction The definition of the fields matches single precision. The double precision bias is 1023. What value can you not represent because of the hidden bit? Certain bit patterns are reserved to represent special values. Of particular importance is the representation for zero (all bits zero). There are also patterns to represent infinity, positive and negative numeric overflow, positive and negative numeric underflow, and not-a-number (abbreviated NaN).

23. Dale Roberts The 32 Bit Single Precision Floating Point Format for IBM 370 Base = 16 Exponent = Excess-64 notation (i.e., compute binary equivalent, then substrate 64) Sign = sign of number (0: positive, 1: negative) Mantissa = normalized fraction (i.e. first digital after ‘.’ is non-zero) Example: What is the value of the following point number? 1 100 0010 1001 0011 1101 0111 1100 0010 Sign = 1  the number is negative Exponent (E) = 100 0010 2 = 66 10 = 2 (substrate 64, because of Excess-64 ) Mantissa (f ) = 1001 0011 1101 0111 1100 0010 = 93D7C2 H   The above floating point number is: x = (sign) f 16 E = - 0.93D7C2 16 2 x = - ( 9 x 16 -1 + 3 x 16 -2 + D x 16 -3 + 7 x 16 -4 + C x 16 -5 + 2 x 16 -6 ) 16 2 = - ( 9 x 16 1 + 3 x 16 0 + 13 x 16 -1 + 7 x 16 -2 + 12 x 16 -3 + 2 x 16 -4 ) = - (144+3 +0.8125+0.02734375 + 0.0029296875 + 0.000030517578125) = - 147.842803955078125 Real Life Example: IBM 370 mantissa 24 bits exponent 7 bits sign 1 bit

24. Dale Roberts Acknowledgements These slides where originally prepared by Dr. Jeffrey Huang, updated by Dale Roberts. IEEE 754 information was obtained from Steve Hollasch http://stevehollasch.com/cgindex/coding/ieeefloat.html. http://stevehollasch.com/cgindex/coding/ieeefloat.html IEEE 754 examples were obtained from Tony Cassandra at St. Edward’s University.