计算机组成原理第二讲 第二章 : 运算方法和运算器 数据与文字的表示方法 (1) 整数的表示方法. 授课老师 : 王浩宇

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1 计算机组成原理第二讲 第二章 : 运算方法和运算器 数据与文字的表示方法 (1) 整数的表示方法 授课老师 : 王浩宇 haoyuwang@bupt.edu.cn 1

2 Today: Bits, Bytes, and Integers Representing information as bits Bit-level manipulations Integers Representation: unsigned and signed Conversion, casting Expanding, truncating Summary 2

3 Everything is bits Each bit is 0 or 1 By encoding/interpreting sets of bits in various ways Computers determine what to do (instructions) and represent and manipulate numbers, sets, strings, etc Why bits? Electronic Implementation Easy to store with bitable elements Reliably transmitted on noisy and inaccurate wires 1.1V 0.9V V 0.0V 3

4 For example, can count in binary Base 2 Number Representation Represent as Represent as [0011] 2 Represent X 10 4 as X

5 Encoding Byte Values Byte = 8 bits Binary to Decimal: 010 to Hexadecimal 0016 to FF16 Base 16 number representation Use characters 0 to 9 and A to F Write FA1D37B16 in C as 0xFA1D37B 0xfa1d37b A B C D E F

6 Example Data Representations C Data Type Typical 32-bit Typical 64-bit x86-64 char short int long float double long double 10/16 pointer

7 Today: Bits, Bytes, and Integers Representing information as bits Bit-level manipulations Integers Representation: unsigned and signed Conversion, casting Expanding, truncating Addition, negation, multiplication, shifting Summary Representations in memory, pointers, strings 7

8 Boolean Algebra Developed by George Boole in 19th Century Algebraic representation of logic And Encode True as 1 and False as 0 A&B = 1 when both A=1 and B=1 Or A B = 1 when either A=1 or B=1 Not ~A = 1 when A=0 Exclusive-Or (Xor) A^B = 1 when either A=1 or B=1, but not both 8

9 General Boolean Algebras Operate on Bit Vectors Operations applied bitwise & ^ ~ All of the Properties of Boolean Algebra Apply 9

10 Bit-Level Operations in C Operations &,, ~, ^ Available in C Apply to any integral data type long, int, short, char, unsigned View arguments as bit vectors Arguments applied bit-wise 10

11 Contrast: Logic Operations in C Contrast to Logical Operators &&,,! View 0 as False Anything nonzero as True Always return 0 or 1 Early termination 11

12 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 left Undefined Behavior Shift amount < 0 or word size Argument x << 3 Log. >> 2 Arith. >> 2 Argument x << 3 Log. >> 2 Arith. >>

13 Today: Bits, Bytes, and Integers Representing information as bits Bit-level manipulations Integers Representation: unsigned and signed Conversion, casting Expanding, truncating Addition, negation, multiplication, shifting Summary Representations in memory, pointers, strings Summary 13

14 Typical ranges for integral data types 14

15 Encoding Integers - Unsigned w 1 B2U(X) x i 2 i i 0 B2U w :{0,1} w -> {0,, 2 w -1} 对于任意整数 x, 当 0 =< x < 2 w 的时候, 存在唯一的二进制序列与其对应 反过来也是一样的, 对于任意一个 w 位的二进制序列, 都存在唯一一个整数 x 满足 0 =< x < 2 w, 与这个二进制序列对应 U2B(x) = B2U -1 (x) 15

16 Encoding Integers - Two s Complement( 补码 ) w 2 B2T(X) x w 1 2 w 1 x i 2 i i 0 Sign Bit TMin = 2 w TMax = 2 w Minus B2T w : {0,1} w -> {-2 w-1,, 2 w-1-1} 对于任意整数 x, 当 -2 w-1 =< x < 2 w-1 的时候, 存在唯一的二进制序列与其对应 反过来, 对于任意一个 w 位的二进制序列, 都存在唯一一个整数 x 满足 -2 w-1 =< x < 2 w-1, 与这个二进制序列对应 T2B(x) = B2T -1 (x) 16

17 Two-complement Encoding Example (Cont.) x = 15213: y = : Weight Sum

18 Values for Different Word Sizes W UMax ,535 4,294,967,295 18,446,744,073,709,551,615 TMax ,767 2,147,483,647 9,223,372,036,854,775,807 TMin ,768-2,147,483,648-9,223,372,036,854,775,808 Observations TMin = TMax + 1 Asymmetric range UMax = 2 * TMax + 1 C Programming #include <limits.h> Declares constants, e.g., ULONG_MAX LONG_MAX LONG_MIN Values platform specific 18

19 有符号数为什么使用补码表示? 原码 (Sign-Magnitude) 最高有效位是符号位, 用来确定剩下的位应该是正权还是负权 比如如果是 8 位二进制 : 0 有两种表示法 [+0] 原 = 0000 ; [-0] 原 = 1000 数据表示范围 整数 : -2 n <X<2 n ( 若数值位 n=3 即 :-8<X<8) 优点 与真值对应关系简单 缺点 参与运算复杂, 需要将数值位与符号位分开考虑 19

20 补码的意义 要将指向 5 点的时钟调整到 3 点整, 应如何处理? 5-2=3 5+10=3(12 自动丢失 12 就是模 ) 20 20

21 补码的意义 继续推导 : 5-2=5+10(MOD 12) 5+(-2)=5+10(MOD 12) -2=10(MOD 12) 结论 : 在模为 12 的情况下,-2 的补码就是 10 一个负数用其补码代替, 同样可以得到正确的运算结果 21 21

22 补码的意义 进一步结论 : 在计算机中, 机器能表示的数据位数是固定的, 其运算都是有模运算 若是 n 位整数 ( 包含符号位 ), 则其模为 2 n ; 若是小数, 则其模为 2 若运算结果超出了计算机所能表示的数值范围, 则只保留它的小于模的低 n-1 位的数值, 超过 n-1 位的高位部分就自动舍弃了 22 22

23 由原码求补码 由原码求补码的简便原则 ( 负数 ) 除符号位以外, 其余各位按位取反, 末位加 1; 从最低位开始, 遇到的第一个 1 以前的各位保持不变, 之后各位取反 例 :[X] 原 = [X] 补 =

24 补码的表示 24

25 Today: Bits, Bytes, and Integers Representing information as bits Bit-level manipulations Integers Representation: unsigned and signed Conversion, casting Expanding, truncating Addition, negation, multiplication, shifting Summary Representations in memory, pointers, strings 25

26 Mapping Between Signed & Unsigned Two s Complement x T2B T2U X B2U Unsigned ux Maintain Same Bit Pattern Unsigned ux U2B U2T X B2T Two s Complement x Maintain Same Bit Pattern Mappings between unsigned and two s complement numbers: Keep bit representations and reinterpret 26

27 Mapping Signed Unsigned Bits Signed Unsigned T2U U2T

28 Mapping Signed Unsigned Bits Signed Unsigned = /

29 Relation between Signed & Unsigned Two s Complement x T2B T2U X B2U Unsigned ux Maintain Same Bit Pattern ux x w Large negative weight becomes Large positive weight 29

30 Relation between Signed & Unsigned WHY? 30

31 Relation between Signed & Unsigned B2U w (x) - B2T w (x) = x w-1 2 w-1 - (-x w-1 2 w-1 ) = x w-1 2 w B2U w (x) = x w-1 2 w + B2T w (x) 令 x 为 T2B w (x), B2U w (T2B w (x)) = x w-1 2 w + B2T w (T2B w (x)) = x w-1 2 w + x T2U w (x) = x w-1 2 w + x 31

32 Conversion Visualized 2 s Comp. Unsigned Ordering Inversion Negative Big Positive TMax UMax UMax 1 TMax + 1 TMax Unsigned Range 2 s Complement Range TMin 32

33 Relation between Signed & Unsigned U2T w = B2T w (U2B w (x)) WHY? 33

34 Relation between Signed & Unsigned U2T w = B2T w (U2B w (x)) B2U w (u) - B2T w (u) = u w-1 2 w-1 - (-u w-1 2 w-1 ) = u w-1 2 w B2T w (u) = B2U w (u) - u w-1 2 w 令 u 为 U2B w (u), B2T w (U2B w (u)) = B2U w (U2B w (u)) - u w-1 2 w = u - u w-1 2 w U2T w (u) = u - u w-1 2 w 34

35 Relation between Signed & Unsigned 35

36 Signed vs. Unsigned in C 36

37 Signed vs. Unsigned in C Constants By default are considered to be signed integers Unsigned if have U as suffix 0U, U Casting Explicit casting between signed & unsigned same as U2T and T2U int tx, ty; unsigned ux, uy; tx = (int) ux; uy = (unsigned) ty; Implicit casting also occurs via assignments and procedure calls tx = ux; uy = ty; 37

38 Casting Surprises Expression Evaluation If there is a mix of unsigned and signed in single expression, signed values implicitly cast to unsigned Including comparison operations <, >, ==, <=, >= Examples for W = 32: TMIN = -2,147,483,648, TMAX = 2,147,483,647 Constant 1 Constant 2 Relation Evaluation 0 0 0U == unsigned < signed U > unsigned > signed U U < unsigned > signed (unsigned) > unsigned U < unsigned (int) U > signed 38

39 Summary Casting Signed Unsigned: Basic Rules Bit pattern is maintained But reinterpreted Can have unexpected effects: adding or subtracting 2 w Expression containing signed and unsigned int int is cast to unsigned!! 39

40 Today: Bits, Bytes, and Integers Representing information as bits Bit-level manipulations Integers Representation: unsigned and signed Conversion, casting Expanding, truncating Addition, negation, multiplication, shifting Summary Representations in memory, pointers, strings 40

41 Sign Extension Task: Given w-bit signed integer x Convert it to w+k-bit integer with same value Rule: Make k copies of sign bit: X = x w 1,, x w 1, x w 1, x w 2,, x 0 k copies of MSB X w X k w 41

42 Sign Extension WHY? X = x w 1,, x w 1, x w 1, x w 2,, x 0 k copies of MSB 42

43 Sign Extension Example short int x = 15213; int ix = (int) x; short int y = ; int iy = (int) y; Decimal Hex Binary x B 6D ix B 6D y C iy FF FF C Converting from smaller to larger integer data type C automatically performs sign extension 43

44 Truncation 44

45 Truncation: Simple Example 45

46 Summary: Expanding, Truncating: Basic Rules Expanding (e.g., short int to int) Unsigned: zeros added Signed: sign extension Both yield expected result Truncating (e.g., unsigned to unsigned short) Unsigned/signed: bits are truncated Result reinterpreted Unsigned: mod operation Signed: similar to mod For small numbers yields expected behavior 46

47 练习题 (1) 47

48 练习题 (2) 48