![Twos Complement
An n-bit, twos complement
number can represent the
range [2 1 , 2 1 1].
Note the asymmetry of this
range about 0 theres one more
negative number than positive
Note what happens when
you overflow
4-bit twos complement range](https://image.slidesharecdn.com/floatingpointrepresentation-230110162052-431721a1/85/Floating_point_representation-pdf-2-320.jpg)
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- 1. 2s Complement and Floating-Point CSE 351 Section 3
- 2. Twos Complement An n-bit, twos complement number can represent the range [2 1 , 2 1 1]. Note the asymmetry of this range about 0 theres one more negative number than positive Note what happens when you overflow 4-bit twos complement range
- 3. Understanding Twos Complement An easier way to find the decimal value of a twos complement number: ~x + 1 = -x We can rewrite this as x = ~(-x - 1), i.e. subtract 1 from the given number, and flip the bits to get the positive portion of the number. Example: 0b11010110 Subtract 1: 0b11010110 - 1 = 0b11010101 Flip the bits: ~0b11010101 = 0b00101010 Convert to decimal: 0b00101010 = (32+8+2)10 = 4210 Multiply by negative one, Answer: -4210.
- 4. Twos Complement: Operations 1. Convert numbers to binary 0xAB = 0b10101011 1710 = 0b00010001 2. Compute x | y 0000 ... 1010 1011 | 0000 ... 0001 0001 -------------------- 0000 ... 1011 1011 3. Compute ~(x | y) (flip the bits) ~ 0000 0000 0000 0000 0000 0000 1011 1011 ----------------------------------------- 1111 1111 1111 1111 1111 1111 0100 0100 int x = 0xAB; int y = 17; int z = 5; int result = ~(x | y) + z; What is the value of result in decimal?
- 5. Twos Complement: Operations 4. Compute ~(x | y) + z You can either convert to decimal before or after adding z 1710 = 0b00010001 Convert 111111111111111111111111010001002 to decimal (hint: -x = ~x + 1) ~111111111111111111111111010001002 = 0000000000000000000000000101110112 ~0000101110112 + 1 = 0000101111002 -x = 0000101111002 x = 1000101111002 int x = 0xAB; int y = 17; int z = 5; int result = ~(x | y) + z; What is the value of result in decimal?
- 6. Floating Point value = (-1)S * M * 2E Numerical Form Sign bit s determines whether number is negative or positive Significand (mantissa) M normally a fractional value in range [1.0, 2.0) Exponent E weights value by a (possibly negative) power of two Representation in Memory MSB s is sign bit s exp field encodes E (but is not equal to E) remember the bias Frac field encodes M (but is not equal to M) s exp mant
- 7. Floating Point value = (-1)S * M * 2E Value: 賊1 Mantissa 2Exponent Bit Fields: (-1)S 1.M 2(E+bias) Bias Read exponent as unsigned, but with bias of (2w-1-1) = 127 Representable exponents roughly 遜 positive and 遜 negative Exponent 0 (Exp = 0) is represented as E = 0b 0111 1111 Why? Floating point arithmetic = easier Somewhat compatible with 2s complement s exp mant
- 8. Floating Point Exponent overflow yields + or - Floats with value +, -, and NaN can be used in operations Result usually still +, -, or NaN; sometimes intuitive, sometimes not Floating point ops do not work like real math, due to rounding! Not associative: (3.14 + 1e100) 1e100 != 3.14 + (1e100 1e100) Not distributive: 100 * (0.1 + 0.2) != 100 * 0.1 + 100 * 0.2 Not cumulative Repeatedly adding a very small number to a large one may do nothing s exp mant
- 9. Floating Point Examples How do you represent -1.5 in floating point? Sign bit: 1 First the integral part of the value: 1 = 0b1 Now compute the decimal: 0.5 = 0b0.1 1.510 = 1.1b Dont need to normalize because its already in scientific notation: 1.1 x 20 Exponent: 0 + 127 = 12710 = 011111112 Mantissa: 10000000000000000000002 -1.510 = 1 01111111 10000000000000000000002 = 0xBFC00000 s exp mant 1 8 23
- 10. Floating Point Examples How do you represent 1.0 in floating point? Sign bit: 0 First the integral part of the value: 1 = 0b1 1.010 => 1.0b Dont need to normalize because its already in scientific notation: 1.0 x 20 Exponent: 0 + 127 = 12710 = 011111112 Mantissa: 000000000000000000000002 1.010 = 0 01111111 000000000000000000000002 = 0x3F800000 s exp mant 1 8 23
- 11. Floating Point Addition (1)s1*M1*2E1 + (-1)s2*M2*2E2 (Assume E1 > E2) Exact Result: (1)s*M*2E Sign s, mantissa M: M = M1 + M2, result of signed align & add Exponent E: E1 Fixing If M 2, shift M right, increment E if M < 1, shift M left k positions, decrement E by k Overflow if E out of range Round M to fit frac precision s exp mant 1 8 23 What is floating point result of 12.3125 + 1.5? 12.312510 = 0 10000010 10001010000000000000000 1.510 = 0 01111111 10000000000000000000000 In scientific notation: (1.1000101 x 23) + (1.1 x 20 ) 1100.0101 x 20 + 1.1 x 20 ------------------ 1101.1101 x 20 Adjusted mantissa: 1.1011101 => M = 1011101 Ans: 0 10000010 10111010000000000000000
- 12. Floating Point Multiplication (1)s1*M1*2E1 * (1)s2*M2*2E2 Exact Result: (1)s*M*2E Sign s: s1 ^ s2 Mantissa M: M1 * M2 Exponent E: E1 + E2 Fixing If M 2, shift M right, increment E If E out of range, overflow Round M to fit frac precision s exp mant 1 8 23 What is floating point result of 1.1* 1.0? 1.110 = 0 01111111 00011001100110011001101 1.010 = 0 01111111 00000000000000000000000 In scientific notation: (1.1 x 20) + (1.0 x 20 ) 1.00011001100110011001101 x 20 * 1.0 x 20 -------------------------------------- 0000000000000000000000000 1000110011001100110011010 Result: 1.000110011001100110011010 x 20 Ans: 0 01111111 00011001100110011001101