Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Number Systems Number of symbols = base of the system Most intuitive -- base 10 (decimal system) counting on fingertips symbols -- 0 1 2 3 4 5 6 7 8 9 Computers use base 2 (binary system) only two symbols - 0 / 1 Many physical reasons for choosing Other useful bases 8 - octal 16 - hexadecimal –1– 52011 Representing a number in base B General expression : A number U3 U2 U1 U0 in base B is effectively U 3 x B 3 + U 2 x B 2 + U 1 x B1 + U 0 x B 0 Examples : 234 in base 10 is 2 x 102 + 3 x 101 + 4 x 100 1011 in base 2 is 1 x 22 + 0 x 22 + 1 x 21 + 1 x 20 = 1 x 101 + 1 x 100 in base 10 –2– 52011 Why Don’t Computers Use Base 10? Implementing Electronically Hard to store Hard to transmit Messy to implement digital logic functions Binary Electronic Implementation Easy to store with bistable elements Reliably transmitted on noisy and inaccurate wires Straightforward implementation of arithmetic functions 0 1 0 3.3V 2.8V 0.5V 0.0V –3– 52011 Octal and hexadecimal Octal uses 0-7 for representing numbers Hexadecimal uses 0-9, a, b, c, d, e and f as digits a = decimal 10, b=decimal 11 and so on They are useful because translation from binary is easy Consider 101100110101 rewrite as 101 100 110 101 and you get 5465 in octal rewrite as 1011 0011 0101 and you get b35 in hex Much more readable in octal and hex than in binary Also much easier to calculate equivalent decimal value It will really pay to learn to interpret hex numbers in this course –4– 52011 Encoding Byte Values Byte = 8 bits Binary 000000002 Decimal: 010 Hexadecimal 0016 to to to 111111112 25510 FF16 Base 16 number representation Use characters ‘0’ to ‘9’ and ‘A’ to ‘F’ Write FA1D37B16 in C as 0xFA1D37B o Or 0xfa1d37b –5– 0 1 2 3 4 5 6 7 8 9 A B C D E F 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 52011 Machine Words Machine Has “Word Size” Nominal size of integer-valued data Including addresses Most current machines are 32 bits (4 bytes) or 64 bits 32 bits Limits addresses to 4GB Becoming too small for memory-intensive applications High-end systems are 64 bits (8 bytes) Potentially address 1.8 X 1019 bytes Machines support multiple data formats Fractions or multiples of word size Always integral number of bytes –6– 52011 Data Representations Sizes of C Objects (in Bytes) C Data Type Compaq Alpha int long int char short float double long double char * 4 8 1 2 4 8 8 8 Typical 32-bit Intel IA32 4 4 1 2 4 8 8 4 4 4 1 2 4 8 10/12 4 o Or any other pointer –7– 52011 Byte Ordering How should bytes within multi-byte word be ordered in memory? Conventions Sun’s, Mac’s are “Big Endian” machines Least significant byte has highest address Alphas, PC’s are “Little Endian” machines Least significant byte has lowest address –8– 52011 Byte Ordering Example Big Endian Least significant byte has highest address Little Endian Least significant byte has lowest address Example Variable x has 4-byte representation 0x01234567 Address given by &x is 0x100 Big Endian 0x100 0x101 0x102 0x103 01 Little Endian 45 67 0x100 0x101 0x102 0x103 67 –9– 23 45 23 01 52011 Reading Byte-Reversed Listings Disassembly Text representation of binary machine code Generated by program that reads the machine code Example Fragment Address 8048365: 8048366: 804836c: Instruction Code 5b 81 c3 ab 12 00 00 83 bb 28 00 00 00 00 Assembly Rendition pop %ebx add $0x12ab,%ebx cmpl $0x0,0x28(%ebx) Deciphering Numbers – 10 – Value: Pad to 4 bytes: Split into bytes: Reverse: 0x12ab 0x000012ab 00 00 12 ab ab 12 00 00 52011 Representing Integers int A = 15213; int B = -15213; long int C = 15213; Linux/Alpha A 6D 3B 00 00 Linux/Alpha B 93 C4 FF FF – 11 – Decimal: 15213 Binary: 0011 1011 0110 1101 3 Hex: B 6 D Sun A Linux C Alpha C Sun C 00 00 3B 6D 6D 3B 00 00 6D 3B 00 00 00 00 00 00 00 00 3B 6D Sun B FF FF C4 93 Two’s complement representation (Covered later) 52011 Alpha P Representing Pointers int B = -15213; int *P = &B; Alpha Address 1 Hex: Binary: Sun P EF FF FB 2C F F F F F C A 0 0001 1111 1111 1111 1111 1111 1100 1010 0000 A0 FC FF FF 01 00 00 00 Sun Address Hex: Binary: E F F F F B 2 C 1110 1111 1111 1111 1111 1011 0010 1100 Linux P D4 F8 FF BF Linux Address Hex: Binary: B F F F F 8 D 4 1011 1111 1111 1111 1111 1000 1101 0100 Different compilers & machines assign different locations to objects – 12 – 52011 Representing Floats Float F = 15213.0; Linux/Alpha F 00 B4 6D 46 Sun F 46 6D B4 00 IEEE Single Precision Floating Point Representation Hex: Binary: 15213: 4 6 6 D B 4 0 0 0100 0110 0110 1101 1011 0100 0000 0000 1110 1101 1011 01 Not same as integer representation, but consistent across machines Can see some relation to integer representation, but not obvious – 13 – 52011 Representing Strings char S[6] = "15213"; Strings in C Represented by array of characters Each character encoded in ASCII format Standard 7-bit encoding of character set Other encodings exist, but uncommon Character “0” has code 0x30 o Digit i has code 0x30+i String should be null-terminated Linux/Alpha S Sun S 31 35 32 31 33 00 31 35 32 31 33 00 Final character = 0 Compatibility Byte ordering not an issue Data are single byte quantities – 14 – Text files generally platform independent 52011 Except for different conventions of line termination character(s)! Boolean Algebra Developed by George Boole in 19th Century Algebraic representation of logic Encode “True” as 1 and “False” as 0 And Not Or A&B = 1 when both A=1 and B=1 & 0 1 0 0 0 1 0 1 ~A = 1 when A=0 ~ 0 1 1 0 – 15 – A|B = 1 when either A=1 or B=1 | 0 1 0 0 1 1 1 1 Exclusive-Or (Xor) A^B = 1 when either A=1 or B=1, but not both ^ 0 1 0 0 1 1 1 0 52011 Application of Boolean Algebra Applied to Digital Systems by Claude Shannon 1937 MIT Master’s Thesis Reason about networks of relay switches Encode closed switch as 1, open switch as 0 A&~B A Connection when ~B A&~B | ~A&B ~A B ~A&B – 16 – = A^B 52011 Algebra Integer Arithmetic Addition is “sum” operation Multiplication is “product” operation – is additive inverse 0 is identity for sum 1 is identity for product Boolean Algebra – 17 – {0,1}, |, &, ~, 0, 1 forms a “Boolean algebra” Or is “sum” operation And is “product” operation ~ is “complement” operation (not additive inverse) 0 is identity for sum 1 is identity for product 52011 Boolean Algebra Integer Ring Commutativity A|B = B|A A&B = B&A Associativity (A | B) | C = A | (B | C) (A & B) & C = A & (B & C) Product distributes over sum A & (B | C) = (A & B) | (A & C) Sum and product identities A|0 = A A&1 = A Zero is product annihilator A&0 = 0 Cancellation of negation ~ (~ A) = A – 18 – A+B = B+A A*B = B*A (A + B) + C = A + (B + C) (A * B) * C = A * (B * C) A * (B + C) = A * B + B * C A+0 = A A*1 =A A*0 = 0 – (– A) = A 52011 Boolean Algebra Integer Ring Boolean: Sum distributes over product A | (B & C) = (A | B) & (A | C) A + (B * C) (A + B) * (B + C) Boolean: Idempotency A|A = A A +AA “A is true” or “A is true” = “A is true” A&A = A Boolean: Absorption A | (A & B) = A A *AA A + (A * B) A “A is true” or “A is true and B is true” = “A is true” A & (A | B) = A Boolean: Laws of Complements A | ~A = 1 A * (A + B) A A + –A 1 “A is true” or “A is false” Ring: Every element has additive inverse A | ~A 0 A + –A = 0 – 19 – 52011 Boolean Ring Properties of & and ^ {0,1}, ^, &, , 0, 1 Identical to integers mod 2 is identity operation: (A) = A A^A=0 Property – 20 – Commutative sum Commutative product Associative sum Associative product Prod. over sum 0 is sum identity 1 is prod. identity 0 is product annihilator Additive inverse Boolean Ring A^B = B^A A&B = B&A (A ^ B) ^ C = A ^ (B ^ C) (A & B) & C = A & (B & C) A & (B ^ C) = (A & B) ^ (B & C) A^0 = A A&1 = A A&0=0 A^A = 0 52011 Relations Between Operations DeMorgan’s Laws Express & in terms of |, and vice-versa A & B = ~(~A | ~B) o A and B are true if and only if neither A nor B is false A | B = ~(~A & ~B) o A or B are true if and only if A and B are not both false Exclusive-Or using Inclusive Or A ^ B = (~A & B) | (A & ~B) o Exactly one of A and B is true A ^ B = (A | B) & ~(A & B) o Either A is true, or B is true, but not both – 21 – 52011 General Boolean Algebras Operate on Bit Vectors Operations applied bitwise 01101001 & 01010101 01000001 01000001 01101001 | 01010101 01111101 01111101 01101001 ^ 01010101 00111100 00111100 ~ 01010101 10101010 10101010 All of the Properties of Boolean Algebra Apply – 22 – 52011 Bit-Level Operations in C Operations &, |, ~, ^ Available in C Apply to any “integral” data type long, int, short, char View arguments as bit vectors Arguments applied bit-wise Examples (Char data type) ~0x41 --> ~010000012 ~0x00 --> ~000000002 0x69 & 0x55 0xBE --> 101111102 0xFF --> 111111112 --> 0x41 011010012 & 010101012 --> 010000012 0x69 | 0x55 --> 0x7D 011010012 | 010101012 --> 011111012 – 23 – 52011 Contrast: Logic Operations in C Contrast to Logical Operators &&, ||, ! View 0 as “False” Anything nonzero as “True” Always return 0 or 1 Early termination Examples (char data type) – 24 – !0x41 --> !0x00 --> !!0x41 --> 0x00 0x01 0x01 0x69 && 0x55 --> 0x01 0x69 || 0x55 --> 0x01 p && *p (avoids null pointer access) 52011 Shift Operations Left Shift: x << y Shift bit-vector x left y positions Throw away extra bits on left Fill with 0’s on right Right Shift: x >> y Shift bit-vector x right y positions Throw away extra bits on right Logical shift Fill with 0’s on left Arithmetic shift Replicate most significant bit on Argument x 01100010 << 3 00010000 Log. >> 2 00011000 Arith. >> 2 00011000 Argument x 10100010 << 3 00010000 Log. >> 2 00101000 Arith. >> 2 11101000 right Useful with two’s complement integer representation – 25 – 52011 Cool Stuff with Xor void funny(int *x, int *y) { *x = *x ^ *y; /* #1 */ *y = *x ^ *y; /* #2 */ *x = *x ^ *y; /* #3 */ } Bitwise Xor is form of addition With extra property that every value is its own additive inverse A^A=0 – 26 – *x *y Begin A B 1 A^B B 2 A^B (A^B)^B = A 3 (A^B)^A = B A End B A 52011 Storing negative integers sign-magnitude of n bits in the word, the most significant bit is sign 1 indicates negative number, 0 a positive number remaining n-1 bits determine the magnitude 0 has two representations The range of valid values is -(2(n-1)-1) to (2(n-1)-1) Examples o 8 bit words can represent -127 to +127 o 16 bit words can represent -32767 to +32767 Advantages Very straightforward Simple to understand Disadvantage different machinery for addition and subtraction multiple representations of zero. – 27 – 52011 One’s complement The most significant bit is effectively sign bit The range of numbers the same as signed magnitude Positive numbers stored as such Negative numbers are bit-wise inverted Example 4 bit storage – range -7 to +7 o -7 = 1000, +7 = 0111, +3 = 0011, -4 = 1011 8 bit storage – range = -127 t0 128 Zero has two representations : o 0000 0000 o 1111 1111 Hardly ever used for actual storage Useful for understanding 2’s compliment and some other operations – 28 – 52011 Two’s Complement Unsigned B2U(X) w1 xi 2 Two’s Complement i B2T(X) xw1 2 i0 w1 w2 xi 2 i i0 short int x = 15213; short int y = -15213; Sign Bit C short 2 bytes long x y Decimal 15213 -15213 Hex 3B 6D C4 93 Binary 00111011 01101101 11000100 10010011 Sign Bit For 2’s complement, most significant bit indicates sign 0 for nonnegative 1 for negative – 29 – 52011 Two’s compliment is effectively one’s compliment with a “1” added to it. It is a number’s additive inverse Number added to its own two’s compliment results in zero Same circuitry can do arithmetic operations for positive and negative numbers The range is (-2n-1) to (2n-1-1) Example : 13 + (-6) 13 in 4 bit binary is 1101 6 is 0110 It’s one’s complement is 1001 Two’s complement is 1010 Add 1101 and 1010 in 4 bits, ignoring the carry The answer is 0111, with carry of 1, which in decimal is 7 – 30 – 52011 Power-of-2 Multiply with Shift Operation u << k gives u * 2k Both signed and unsigned Operands: w bits True Product: w+k bits * u · 2k Discard k bits: w bits u k • • • 2k 0 ••• 0 1 0 ••• 0 0 • • • UMultw(u , 2k) ••• 0 ••• 0 0 0 ••• 0 0 TMultw(u , 2k) Examples u << 3 == u << 5 - u << 3 == Most machines shift and add much faster than multiply – 31 – u * 8 u * 24 52011 Unsigned Power-of-2 Divide with Shift Quotient of Unsigned by Power of 2 u >> k gives u / 2k Uses logical shift k u Operands: Division: Result: x x >> 1 x >> 4 x >> 8 – 32 – / 2k ••• ••• 0 ••• 0 1 0 ••• 0 0 u / 2k 0 ••• ••• u / 2k 0 ••• ••• Division 15213 7606.5 950.8125 59.4257813 Binary Point Computed 15213 7606 950 59 Hex 3B 6D 1D B6 03 B6 00 3B . ••• Binary 00111011 01101101 00011101 10110110 00000011 10110110 00000000 00111011 52011 Signed Power-of-2 Divide with Shift Quotient of Signed by Power of 2 x >> k gives x / 2k Uses arithmetic shift Rounds wrong direction when u k< 0 x Operands: / x / 2k Division: Result: y y >> 1 y >> 4 y >> 8 – 33 – 2k RoundDown(x / 2k) Division -15213 -7606.5 -950.8125 -59.4257813 ••• ••• Binary Point 0 ••• 0 1 0 ••• 0 0 0 ••• ••• 0 ••• ••• Computed -15213 -7607 -951 -60 Hex C4 93 E2 49 FC 49 FF C4 . ••• Binary 11000100 10010011 11100010 01001001 11111100 01001001 11111111 11000100 52011 Correct Power-of-2 Divide Quotient of Negative Number by Power of 2 Want x / 2k (Round Toward 0) Compute as (x+2k-1)/ 2k In C: (x + (1<<k)-1) >> k Biases dividend toward 0 k Case 1: No rounding u Dividend: +2k +–1 1 / 2k u / 2k 0 ••• 0 0 0 ••• 0 0 1 ••• 1 1 1 Divisor: ••• ••• 1 ••• 1 1 Binary Point 0 ••• 0 1 0 ••• 0 0 0 ••• 1 1 1 1 ••• . 1 ••• 1 1 Biasing has no effect – 34 – 52011 Correct Power-of-2 Divide (Cont.) Case 2: Rounding k x Dividend: +2k +–1 1 ••• ••• 0 ••• 0 0 1 ••• 1 1 1 ••• ••• Incremented by 1 Divisor: / 2k x / 2k 0 ••• 0 1 0 ••• 0 0 0 ••• 1 1 1 1 Biasing adds 1 to final result – 35 – Binary Point ••• . ••• Incremented by 1 52011 Fractional Binary Numbers 2i 2i–1 4 2 1 ••• bi bi–1 ••• b2 b1 b0 . b–1 b–2 b–3 1/2 1/4 1/8 ••• b–j ••• 2–j Representation Bits to right of “binary point” represent fractional powers of 2 i Represents rational number: k bk 2 k j – 36 – 52011 Frac. Binary Number Examples Value 5-3/4 2-7/8 63/64 Representation 101.112 10.1112 0.1111112 Observations Divide by 2 by shifting right Multiply by 2 by shifting left Numbers of form 0.111111…2 just below 1.0 1/2 + 1/4 + 1/8 + … + 1/2i + … 1.0 – 37 – 52011 Representable Numbers Limitation Can only exactly represent numbers of the form x/2k Other numbers have repeating bit representations Value 1/3 1/5 1/10 – 38 – Representation 0.0101010101[01]…2 0.001100110011[0011]…2 0.0001100110011[0011]…2 52011