Numeric/Instruction Representation

1. Numeric/Instruction Representation Define the terms least significant bit and most significant bit. Explain how unsigned integer numbers are represented in memory Understand the limitations of using sign and magnitude to represent signed integer numbers Explain how signed integer numbers are represented in memory using twos complement notation. Convert twos complement numbers into decimal. Convert decimal numbers into twos complement format. Explain the concept of sign extension Convert a binary number into hexadecimal Explain how instructions are represented in memory as machine code

2. Numeric values can be represented in any base. People commonly use base 10 (decimal). Given a decimal (base 10) number, convert it to binary: • 51ten= ?two • -7ten= ?two CS2710 Computer Organization

3. Given a binary number, convert it to decimal format • 1 1011two= ?ten CS2710 Computer Organization

4. Processors use each bit of a word to represent a binary digit of a numeric value • Consider the integer value 305,419,896 represented with a MIPS 32-bit word:00010010001101000101011001111000 In this representation, each bit represents the value xi*2i, where iis the bit number, and x is either 0 or 1. Bit 0(LSB) Bit 31 (MSB) CS2710 Computer Organization

5. Signed numbers: How do we represent negative values (+/-)? • First approach • Add a sign bit to the number • 0 indicates positive • 1 indicates negative • Example: -11ten= ?two CS2710 Computer Organization

6. Dual zeros: the problem with sign and magnitude • 1 00000000 == 0 00000000 Sign bit Sign bit Problem: Two zero values. Not good. Sign bits were used in early computers, but were soon abandoned. No modern computers use sign bits! CS2710 Computer Organization

7. Solution: 2’s complement numbers • Leading 0’s mean positive, leading 1’s mean negative • Example for 32 bits: 1111 1111 1111 1111 1111 1111 1111 11002= –1×231+ 1×230 + … + 1×22 +0×21 +0×20= –2,147,483,648 + 2,147,483,644 = –410 • 32 bits can represent the values –2,147,483,648 to +2,147,483,647 Range: –2n – 1 to +2n – 1 – 1 CS2710 Computer Organization

8. Shortcut: Converting a number into 2’s complement • Convert the absolute value of the number into a binary number • Complement the bits (1-> 0, 0->1) • Add 1 to the value Example: Convert -51 (decimal) to binary using 2’s complement. CS2710 Computer Organization

9. Sign extension • Representing a number using more bits • Preserve the numeric value • Replicate the sign bit to the left • c.f. unsigned values: extend with 0s • Examples: 8-bit to 16-bit • +2: 0000 0010 => 0000 00000000 0010 • –2: 1111 1110 => 1111 11111111 1110 CS2710 Computer Organization

10. Example • Convert -51 to a 16 bit signed binary (base 2) representation based on the 2’s complement value from before CS2710 Computer Organization

11. Converting binary to hex (CE1900 review) • Hexadecimal – Base 16 • Group binary digits into sets of 4 • Convert each group into a hex digit CS2710 Computer Organization

12. Example • Convert the number 10100101 to hex • Convert the number 1100100 to hex • What decimal number is this? CS2710 Computer Organization

13. Representing Instructions • Instructions are represented in binary • Called machine code • MIPS instructions (e.g. add $t0, $s1, $s2) • Encoded as 32-bit words • Small number of formats • Regularity in the pattern • Register numbers $t0 – $t7 are registers8 – 15 $t8 – $t9 are registers24 – 25 $s0 – $s7 are registers16 – 23 §2.5 Representing Instructions in the Computer

14. op rs rt rd shamt funct 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits MIPS R[egister]-format Instructions • Instruction fields • op: operation code (opcode) • rs: first source register number • rt: second source register number • rd: destination register number • shamt: shift amount (00000 for now) • funct: function code (extends opcode) • For add, $t0, $s1, $s2 (see the green card in the textbook) • op = 0 • rs = $s1, register 17 • rt = $s2, register 18 • rd = $t0, register 8 • shamt = 0 • funct = 32

15. op 0 000000 rs 17 10001 18 rt 10010 01000 8 rd 00000 shamt 0 32 100000 funct 6 bits 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits 5 bits 5 bits 5 bits 5 bits 5 bits 5 bits 5 bits 5 bits 6 bits 6 bits 6 bits MIPS R-format for the add instruction • For add, $t0, $s1, $s2 • op = 0 (for add) • rs = $s1, register 17 • rt = $s2, register 18 • rd = $t0, register 8 • shamt = 0 (for add) • funct = 32 (for add) add instruction syntax:add rd, rs, rt (See the green card in the textbook) Q1: why 5 bits for register values??? Q2: What is the hex equivalent?

16. op 0010 00 rs 10 001 rt 0 1000 Constant or address 0000 0000 0110 0100 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits 16 bits 16 bits MIPS I[mmediate]-format Instructions • Instruction fields • op: operation code (opcode) • rs: first source register number • rt: second source register number • Last field: constant value or 16-bit base address offset • For addi, $t0, $s1, 100 # t0 = s1 + 100 • op = 8 • rt = $t0, register 8 • rs = $s1, register 17 • Constant =100 addi instruction syntax:addirt, rs, value (sign-extended) (See the green card in the textbook)

17. op 1000 11 rs 01 001 rt 0 1000 Constant or address 0000 0100 1011 0000 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits 16 bits 16 bits MIPS I[mmediate]-format Instructions • Instruction fields • op: operation code (opcode) • rs: first source register number • rt: second source register number • Last field: constant value or 16-bit base address offset • For lw $t0, 1200($t1) #load value from memory ref’d by t2 into t1 • op = 35 • rt = $t0, register 8 • rs = $t1, register 9 • Address offset=1200 (bytes) lwinstruction syntax:lwrt, (offset)rs, (See the green card in the textbook)