CS/COE0447 Computer Organization & Assembly Language Chapter 2 Part 1

Topics to cover in Chapter 2 MIPS operations and operands MIPS registers Memory view Instruction encoding Arithmetic operations Logic operations Memory transfer operations

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

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

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

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

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)

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

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

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)

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

R-Format Instructions (REMINDER) Define “fields” of the following number of bits each: 6 + 5 + 5 + 5 + 5 + 6 = 32 6 5 For simplicity, each field has a name: opcode rs rt rd funct shamt

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

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: 012A 4020hex decimal representation: 19,546,144ten

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

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}

I-Format Example MIPS Instruction: addi $8,$9,7 Decimal number per field representation: Binary number per field representation:

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: 001000 01001 01000 0000000000000111 hex hex representation: 0x21280007

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}

Executing the addi instruction 00000023 +00000007 0000002A addi $8,$9,7 $8 $9 0x00000023 Suppose  Immediate = 0x0007 (16 bits; 4 hex digits) SignExtImm = 0x00000007 (32 bits; 8 hex digits) This will be more interesting when we get to negative numbers.

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%

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%

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

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

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

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'b1001 // a 4-bit binary number 16'h704f // a 16-bit hex number 1b‘0 // a 1-bit binary number .text lui $t1,0x7F40 addi $t2,$t1,0x777 andi $t3,$t2,0x5555 In class

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 1010101001010101 0000000000000000 $t0 1010101001010101 1100110000110011 $t0

Loading a memory address .data places values in memory starting at 0x10010000. 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 $t0,0x1001008 is a pseudo instruction – not implemented in the hardware lui $1,4097 la $t0,0x10010008 ori $8,$1,8 lw $t1,0($t0)

A program Get sample1.asm from the schedule Load it into the simulator Figure out the memory contents, labels Trace through the code

.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