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**CS/COE0447 Computer Organization & Assembly Language**

Chapter 2 Part 1

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**Topics to cover in Chapter 2**

MIPS operations and operands MIPS registers Memory view Instruction encoding Arithmetic operations Logic operations Memory transfer operations

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**MIPS Operations/Operands**

“Operation” (instruction) “Operand” MIPS operations Arithmetic operations (integer/floating-point) (add, sub,…) Logical operations (and, or,…) Shift operations (shift a certain number of bits to the left or right) Compare operations (do something if one operand is less than another,…) Load/stores to transfer data from/to memory Branch/jump operations System control operations/coprocessor operations MIPS operands General-purpose registers Fixed registers, e.g., HI/LO registers Memory location Immediate value

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MIPS Arithmetic rd rs rt <op> <rdestination> <rsource1> <rsource2> All arithmetic instructions have 3 operands Operand order is fixed: destination first 32 registers (page 2 of green card) Examples add $t0, $s0, $s2 # $t0 = $s0 + $s2 sub $s0, $t0, $t1 # $s0 = $t0 – $t1

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**MIPS Registers $zero r0 r16 $s0 HI $at r1 r17 $s1 LO $v0 r2 r18 $s2**

32 bits 32 bits 32 bits $zero r0 r16 $s0 HI $at r1 r17 $s1 LO $v0 r2 r18 $s2 $v1 r3 r19 $s3 $a0 r4 r20 $s4 $a1 r5 r21 $s5 $a2 r6 r22 $s6 $a3 r7 r23 $s7 $t0 r8 r24 $t8 $t1 r9 r25 $t9 $t2 r10 r26 $k0 $t3 r11 r27 $k1 $t4 r12 r28 $gp $t5 r13 r29 $sp $t6 r14 r30 $fp $t7 r15 r31 $ra PC General-Purpose Registers Special-Purpose Registers

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**General-Purpose Registers**

GPR: all can be used as operands in instructions Still, conventions and limitations exist to keep GPRs from being used arbitrarily r0, termed $zero, always has a value “0” r31, termed $ra (return address), is reserved for subroutine call/return Etc. (we’ll see otherc conventsion/limitations later) Register usage and related software conventions are summarized in “application binary interface” (ABI), which is important when writing system software such as an assembler and a compiler

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**Instruction Encoding Instructions are encoded in binary numbers**

Assembler translates assembly programs into binary numbers Machine decodes binary numbers to figure out what the instruction is MIPS has “fixed” 32-bit instruction encoding MIPS has several instruction formats R-format: arithmetic instructions I-format: transfer/branch/immediate format J-format: jump instruction format (FI/FR-format: floating-point instruction format)(later chapter)

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**MIPS Instruction Formats**

Name Fields Comments 6 bits 5 bits All MIPS instructions 32 bits R-format op rs rt rd shamt funct Arithmetic/logic instruction format I-format address/immediate Transfer, branch, immediate format J-format target address Jump instruction format

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**R-Format Instructions**

Define “fields” of the following number of bits each: = 32 6 5 For simplicity, each field has a name: opcode rs rt rd funct shamt For shift instructions: “shift amount”

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**R-Format Example MIPS Instruction:**

add $8,$9,$10 Decimal number per field representation: Binary number per field representation: hex representation: decimal representation: On Green Card: Format in column 1, opcodes in column 3 (Let’s look and then come back)

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**M I P S Reference Data: CORE INSTRUCTION SET**

NAME MNE-MON-IC FOR-MAT OPERATION (in Verilog) OPCODE/ FUNCT (hex) Add add R R[rd] = R[rs] + R[rt] (1) 0 / 20hex Add Immediate addi I R[rt] = R[rs] + SignExtImm (1)(2) 8hex Branch On Equal beq if(R[rs]==R[rt]) PC=PC+4+ BranchAddr (4) 4hex (1) May cause overflow exception (2) SignExtImm = { 16{immediate[15]}, immediate } (3) ZeroExtImm = { 16{1b’0}, immediate } (4) BranchAddr = { 14{immediate[15]}, immediate, 2’b0} Later

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**R-Format Instructions (REMINDER)**

Define “fields” of the following number of bits each: = 32 6 5 For simplicity, each field has a name: opcode rs rt rd funct shamt

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**R-Format Example MIPS Instruction: Now let’s fill this in**

add $8,$9,$10 Decimal number per field representation: Binary number per field representation: Now let’s fill this in

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**R-Format Example MIPS Instruction: 9 10 8 32 000000 01001 01010 01000**

add $8,$9,$10 Decimal number per field representation: 9 10 8 32 Binary number per field representation: hex 000000 01001 01010 01000 100000 00000 hex representation: A 4020hex decimal representation: ,546,144ten

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**I-Format Instructions**

Define “fields” of the following number of bits each: = 32 6 5 5 16 For simplicity, each field has a name: opcode rs rt immediate Let’s do an example using addi

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**M I P S Reference Data: CORE INSTRUCTION SET**

NAME MNE-MON-IC FOR-MAT OPERATION (in Verilog) OPCODE/ FUNCT (hex) Add add R R[rd] = R[rs] + R[rt] (1) 0 / 20hex Add Immediate addi I R[rt] = R[rs] + SignExtImm (1)(2) 8hex Branch On Equal beq if(R[rs]==R[rt]) PC=PC+4+ BranchAddr (4) 4hex (1) May cause overflow exception (2) SignExtImm = { 16{immediate[15]}, immediate } (3) ZeroExtImm = { 16{1b’0}, immediate } (4) BranchAddr = { 14{immediate[15]}, immediate, 2’b0}

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**I-Format Example MIPS Instruction: addi $8,$9,7**

Decimal number per field representation: Binary number per field representation:

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**I-Format Example MIPS Instruction: 001000 01001 01000 0000000000000111**

addi $8,$9,7 Decimal number per field representation: 8 9 8 7 Binary number per field representation: hex hex representation: 0x

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**M I P S Reference Data: CORE INSTRUCTION SET**

NAME MNE-MON-IC FOR-MAT OPERATION (in Verilog) OPCODE/ FUNCT (hex) Add add R R[rd] = R[rs] + R[rt] (1) 0 / 20hex Add Immediate addi I R[rt] = R[rs] + SignExtImm (1)(2) 8hex Branch On Equal beq if(R[rs]==R[rt]) PC=PC+4+ BranchAddr (4) 4hex (1) May cause overflow exception (2) SignExtImm = { 16{immediate[15]}, immediate } (3) ZeroExtImm = { 16{1b’0}, immediate } (4) BranchAddr = { 14{immediate[15]}, immediate, 2’b0}

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**Executing the addi instruction**

A addi $8,$9,7 $8 $9 0x Suppose Immediate = 0x0007 (16 bits; 4 hex digits) SignExtImm = 0x (32 bits; 8 hex digits) This will be more interesting when we get to negative numbers.

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**Exercise Which instruction has same representation as 35ten?**

A. add $0, $0, $0 B. subu $s0,$s0,$s0 C. lw $0, 0($0) D. addi $0, $0, 35 E. subu $0, $0, $0 F. Trick question! Instructions are not numbers. Use Green Card to Answer 1 PERSON VOTE: Pink: 40%, Reed: 30%, Yellow: 20%, Blue: 20% Afterwards: Pink 90%

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**Exercise Which instruction has same representation as 35ten? 32 16 35**

A. add $0, $0, $0 B. subu $s0,$s0,$s0 C. lw $0, 0($0) D. addi $0, $0, 35 E. subu $0, $0, $0 F. Trick question! Instructions are not numbers. Registers numbers and names: 0: $0, 8: $t0, 9:$t1, …,16: $s0, 17: $s1, …, Opcodes and function fields add: opcode = 0, function field = 32 subu: opcode = 0, function field = 35 addi: opcode = 8 lw: opcode = 35 35 32 8 16 1 PERSON VOTE: Pink: 40%, Reed: 30%, Yellow: 20%, Blue: 20% Afterwards: Pink 90%

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**Which instruction bit pattern = number 35?**

A. add $0, $0, $0 B. subu $s0,$s0,$s0 C. lw $0, 0($0) D. addi $0, $0, 35 E. subu $0, $0, $0 F. 35 32 8 16

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**Logic instruction format**

Logic Instructions Name Fields Comments R-format op rs rt rd shamt funct Logic instruction format Bit-wise logic operations <op> <rdestination> <rsource1> <rsource2> Examples and $t0, $s0, $s2 # $t0 = $s0 ^ $s2 or $s0, $t0, $t1 # $s0 = $t0 | $t1 nor $s0, $t0, $t1 # $s0 = ~($t0 | $t1) xor $s0, $t0, $t1 # $s0 = $t0 ^ $t1

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**Logic Instructions: Example**

.text addi $t0,$0,0x32 addi $t1,$0,0x777 and $t2,$t1,$t0 Answer in class; also, replace and by or Next: andi and ori

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**Andi and Ori andi I R[rt] & ZeroExtImm (3)**

lui I R[rt] = {immediate,16’b0} (3) ZeroExtImm = {16{1’b0},immediate} In Verilog: 4'b // a 4-bit binary number 16'h704f // a 16-bit hex number 1b‘ // a 1-bit binary number .text lui $t1,0x7F40 addi $t2,$t1,0x777 andi $t3,$t2,0x5555 In class

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**Long Immediates (review)**

Sometimes we need a long immediate, e.g., 32 bits MIPS requires that we use two instructions lui $t0, 0xaa55 Then we get lower-order 16 bits ori $t0, $t0, 0xcc33 $t0 $t0

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**Loading a memory address**

.data places values in memory starting at 0x So, 32 bits are needed to specify a memory address. Format I has a 16 bit field and Format J has a 26 bit field…neither is long enough. la $t0,0x is a pseudo instruction – not implemented in the hardware lui $1, la $t0,0x ori $8,$1,8 lw $t1,0($t0)

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**A program Get sample1.asm from the schedule Load it into the simulator**

Figure out the memory contents, labels Trace through the code

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.data # sample1.asm a: .word 3,4 c: .word 5,6 .text la $t0,c # address of c la $t1,k # address of k lw $s0,0($t0) # load c[0] lw $s1,4($t1) # load k[1] slt $s3,$s0,$s1 # if c[0] < k[1], $s3 = 1, else $s3 = 0 beq $s3,$0,notless # if c[0] < k[1] swap their values sw $s0,4($t1) sw $s1,0($t0) notless: .data k: word 0xf,0x11,0x12

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