COMP Overview of Tutorial #2

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1 COMP 1402 Winter 2008 Tutorial #2 Overview of Tutorial #2 Number representation basics Binary conversions Octal conversions Hexadecimal conversions Signed numbers (signed magnitude, one s and two s complement, Excess-M) Float conversions 1

2 Number representation basics In binary each digit (or bit) has 2 possible values: 0 or 1. Ex: ( ) 2 In octal each digit has 8 possible values: 0, 1, 2 7. Ex: (4127) 8 In hexadecimal (or hex for short) each digit has 16 possible values: 0, 1, 2 9, then A, B, C, D, E and F (representing 10, 11, 12, 13,14 and 15 in decimal respectively). Ex: (1AF6) 16 or 1AF6h Arbitrary base b: Bases d n d n-1 d n-2... d 2 d 1 d 0 To calculate the value: d n *b n + d n-1 *b n d 1 *b 1 + d 0 *b 0 Decimals (base 10): How do we interpret the number 2510? = 2* * * *10 0 2

3 Bases (2) Given a number d n d n-1 d n-2... d 2 d 1 d 0 In binary (base 2) value = d n *2 n + d n-1 *2 n d 1 *2 1 + d 0 *2 0 In octal (base 8) value = d n *8 n + d n-1 *8 n d 1 *8 1 + d 0 *8 0 In hexadecimal (base 16) value = d n *16 n + d n-1 *16 n d 1 * d 0 *16 0 Fractions in bases Arbitrary base fractions: 0. d -1 d -2 d -3 value = d -1 *b -1 + d -2 *b -2 + d -3 *b -3 In decimal (base 10): is equal to 5* * *10-3 In binary (base 2): is equal to 1* * * *2-4 = 1* * * * =

4 Binary: Most/Least Significant bits Most significant bit (MSB) is the leftmost bit: Ex: It is the bit of highest value (128 in the example above) but can also be used as the sign of the number (as we ll see later). Least significant bit (LSB) is the rightmost bit: Ex: Has the least value of all the bits (0 or 1). Binary to decimal To convert binary to decimal we must add the digits weighed by exponents of 2 used in the binary number as seen previously. value = d n *2 n + d n-1 *2 n d 1 *2 1 + d 0 *2 0 Ex: convert (11010) 2 to decimal. This is equal to 1*2 4 +1* * *2 1 +0* = 26 (11010) 2 = (26) 10 4

5 Decimal to binary Converting a number from base 10 to base 2. Let s look again at the value of a binary number: d n *2 n + d n-1 *2 n d 1 *2 1 + d 0 *2 0 We need to fill in the d 0 to d n to build the binary number as d n d n-1 d n-2... d 2 d 1 d 0 Different ways to solve this Start with the largest power of 2 in the decimal number, then move down (slow) Algorithm using the mod approach (easier) Decimal to binary (2) Decimal to binary algorithm: Q = decimal number While Q is not equal to 0 do the following: Binary digit = Q mod 2 Q = Q / 2 (quotient) End While Let s try an example 5

6 Decimal to binary (3) Example: convert (134) 10 to binary. 134 mod 2 = 0 67 mod 2 = 1 33 mod 2 = 1 16 mod 2 = 0 8 mod 2 = 0 4 mod 2 = 0 2 mod 2 = 0 1 mod 2 = 1 We then read the numbers from bottom up: So (134) 10 = ( ) 2 Decimal fraction to binary Convert both sides of the period separately. We ve seen how to do the left side, now the fraction side. Instead of mod, we multiply by two (*2). When we go over 1.0, subtract 1 for next round. Ex: Convert ( ) 10 to binary * 2 = * 2 = * 2 = * 2 = * 2 = 1.0 (stop at 1.0) Then we read the numbers top to bottom: Therefore ( ) 10 =

7 Octal to binary Since every octal digit can take 8 values we can convert each digit to binary using 3 bits. Ex: convert (7213) 8 to binary. We ll convert each digit separately: Octal: Binary: Therefore: (7213) 8 = ( ) 2 Binary to octal Group the binary number into groups of 3 bits, starting from the right, then convert each group into their octal value. Ex: convert ( ) 2 to octal Therefore: ( ) 2 = (32635) 8 7

8 Hexadecimal to binary Since every digit of hex has 16 possible values, we represent each digit using 4 binary bits. Ex: convert (5DE9) 16 to binary. convert each digit into four binary bits: 5 D E Therefore (5DE9) 16 = ( ) 2 Binary to hexadecimal Group the binary number into groups of 4 bits, starting from the right, then convert each group into their hex value. Ex: ( ) 2 to hex F 7 Therefore ( ) 2 = (125F7) 16 8

9 Octal to decimal Count the digits weighed by exponents of 8, as seen previously: value = d n *8 n + d n-1 *8 n d 1 *8 1 + d 0 *8 0 Ex: convert (314) 8 to decimal. = 3* * *8 0 = 3*64 + 1*8 + 4*1 = (314) 8 = (204) 10 Hexadecimal to decimal Count the digits weighed by exponents of 16, as seen previously: value = d n *16 n + d n-1 *16 n d 1 * d 0 *16 0 Ex: convert (4A1C) 16 to decimal. = 4* * * *16 0 = 4* * * *1 = (4A1C) 16 =

10 Part I of the exercises then correction Signed Magnitude One s Complement Two s Complement Excess-M (Bias) Signed Numbers 10

11 Signed Magnitude MSB is 1 if the number is negative (-), 0 if the number is positive (+). The rest of the number is converted to binary like we ve seen before. What happens with numbers smaller than -127? overflow Example: (-14) 10 : 14 in binary is: in signed magnitude is: One s complement MSB is the sign bit 1 is negative 0 is positive Rules to change sign: Flip all the bits (change 0 s to 1 s and 1 s to 0 s) Two zeroes: and Harder to know the value of a negative number (have to be complemented first) 11

12 One s Complement (decimal to binary) First, convert the decimal to binary If the number is negative, flip every bit (0 become 1, 1 becomes 0). Positive numbers are converted to binary without change. Example: (-44) 10 in one s complement 44 in binary is: in one s complement is: Example: (38) 10 in one s complement 38 in binary is: in one s complement is: (same) One s Complement (binary to decimal) If the MSB is 1, we must flip every bit before converting to decimal. Ex: convert ( ) 2 from one s complement to decimal. Flip the bits since M.S.B. is a 1 : Then find the decimal value: = 99 and we add the sign= (-99) 10 If the number is positive (MSB is 0 ) we convert to decimal without any other changes. 12

13 Two s complement MSB is the sign bit 1 is negative 0 is positive Rules to change sign: Take one s complement, then add 1. Only one zero: Have to take the two s complement of negatives in order to know the value. Two s Complement (decimal to binary) Convert the decimal to binary, then to one s complement, then add 1. Ex: convert (-57) 10 to two s complement. 57 in binary is: in one s complement: add 1: + 1 in two s complement:

14 Two s Complement (binary to decimal) If the MSB is 1, take one s complement (flip every bit) then add 1. Ex: convert ( ) 2 from two s complement to decimal. Take one s complement : add 1: and convert to decimal = = 52 so the answer is (-52) 10 Excess-M (Bias) 8-bit version has 256 values (-127 to +128). Used to store the exponent in the float rep. For floating point, we ll use Excess-127. Ex: Convert 32 to Excess-127 = = 159 or To convert Excess-127 back to decimal: convert to decimal then subtract 127. Ex: Convert in Excess-127 to decimal = = (79) 10 14

15 Addition in One s complement First convert both decimals in one s complement then add. If there is a carry, add it to the right. Ex: add (-15) 10 and (+69) 10 in one s complement. -15 in one s complement: in one s complement: add: add carry to right: 1 answer: or +54 Addition in Two s complement First convert both decimals in two s complement, then add. If there is a carry, discard it. Ex: add (-22) 10 and (19) 10 in two s complement -22 in two s complement: in two s complement: add: Since MSB is 1, take two s complement to get the answer: or -3 15

16 Float representation Used to represent fractional numbers Overall sign Mantissa (fraction part) Exponent Base In C: 32-bit float (4 bytes) Sign Exponent Mantissa 1 bit 8 bits 23 bits Float to decimal Convert from float (1-bit sign, 8-bit exponent and 23-bit mantissa) to decimal. Sign: - since the bit sign is 1 Exponent: to decimal = 3, then subtract 127 to convert from Excess-127 to true value: = Mantissa: 11 followed by zeroes. We supply the leading 1 (always). So the number is -(1.11) 2 * = *

17 Decimal to float Convert (3.6875) 10 to float (1-bit sign, 8-bit exponent and 23-bit mantissa). Sign: 0 (since the number is positive) To figure out the mantissa and the exponent, we must first convert the decimal to binary. 3 to binary = to binary: * 2 = * 2 = * 2 = * 2 = 1.0 So in binary is (continued next slide) Decimal to float (2) Floats must always be in the format 1.x so we must move the period to have 1.x in our binary number: (we have to shift the period one place to the left). So = * 2 1 We now have the mantissa: 11011(followed by zeroes) And the exponent is 1, converted to Excess-127 = 128. In binary 128 = Answer: Convert back to decimal to verify the answer. 17

18 Part II of the exercises then correction 18

CHAPTER V NUMBER SYSTEMS AND ARITHMETIC

CHAPTER V-1 CHAPTER V CHAPTER V NUMBER SYSTEMS AND ARITHMETIC CHAPTER V-2 NUMBER SYSTEMS RADIX-R REPRESENTATION Decimal number expansion 73625 10 = ( 7 10 4 ) + ( 3 10 3 ) + ( 6 10 2 ) + ( 2 10 1 ) +(