MIPS R-format Instructions. Representing Instructions. Hexadecimal. R-format Example. MIPS I-format Example. MIPS I-format Instructions

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Representing Instructions Instructions are encoded in binary Called machine code MIPS instructions Encoded as 32-bit instruction words Small number of formats encoding operation code (opcode), register numbers, Regularity! Register numbers $t0 $t7 are reg s 8 15 $t8 $t9 are reg s 24 25 $s0 $s7 are reg s 16 23 2.5 Representing Instructions in the Computer MIPS R-format Instructions op rs rt rd shamt funct 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits 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) Chapter 2 Instructions: Language of the Computer 67 Chapter 2 Instructions: Language of the Computer 68 R-format Example op rs rt rd shamt funct 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits add $t0, $s1, $s2 special $s1 $s2 $t0 0 add 0 17 18 8 0 32 000000 10001 10010 01000 00000 100000 00000010001100100100000000100000 2 = 02324020 16 Hexadecimal Base 16 Compact representation of bit strings 4 bits per hex digit 0 0000 4 0100 8 1000 c 1100 1 0001 5 0101 9 1001 d 1101 2 0010 6 0110 a 1010 e 1110 3 0011 7 0111 b 1011 f 1111 Example: eca8 6420 1110 1100 1010 1000 0110 0100 0010 0000 Chapter 2 Instructions: Language of the Computer 69 Chapter 2 Instructions: Language of the Computer 70 MIPS I-format Instructions MIPS I-format Example op rs rt constant or address Immediate arithmetic and load/store instructions rt: destination or source register number Constant: 2 15 to +2 15 1 Address: offset added to base address in rs Design Principle 4: Good design demands good compromises Different formats complicate decoding, but allow 32-bit instructions uniformly Keep formats as similar as possible op rs rt constant or address lw $t0, 32($s3) # Temporary reg $t0 gets A[8] lw $s3 $t0 address 35 19 8 32 100011 10011 01000 0000000000100000 Chapter 2 Instructions: Language of the Computer 71 Chapter 2 Instructions: Language of the Computer 72 1

Stored Program Computers The BIG Picture Instructions represented in binary, just like data Instructions and data stored in memory Programs can operate on programs e.g., compilers, linkers, Binary compatibility allows compiled programs to work on different computers Standardized ISAs Logical Operations Instructions for bitwise manipulation Operation C Java MIPS Shift left << << sll Shift right >> >>> srl Bitwise AND & & and, andi Bitwise OR or, ori Bitwise NOT ~ ~ nor Useful for extracting and inserting groups of bits in a word 2.6 Logical Operations Chapter 2 Instructions: Language of the Computer 73 Chapter 2 Instructions: Language of the Computer 74 Shift Operations op rs rt rd shamt funct 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits shamt: how many positions to shift Shift left logical Shift left and fill with 0 bits sll by i bits multiplies by 2 i Shift right logical Shift right and fill with 0 bits srl by i bits divides by 2 i (unsigned only) AND Operations Useful to mask bits in a word Select some bits, clear others to 0 and $t0, $t1, $t2 $t2 0000 0000 0000 0000 0000 1101 1100 0000 $t1 0000 0000 0000 0000 0011 1100 0000 0000 $t0 0000 0000 0000 0000 0000 1100 0000 0000 Chapter 2 Instructions: Language of the Computer 75 Chapter 2 Instructions: Language of the Computer 76 OR Operations Useful to include bits in a word Set some bits to 1, leave others unchanged or $t0, $t1, $t2 $t2 0000 0000 0000 0000 0000 1101 1100 0000 $t1 0000 0000 0000 0000 0011 1100 0000 0000 NOT Operations Useful to invert bits in a word Change 0 to 1, and 1 to 0 MIPS has NOR 3-operand instruction a NOR b == NOT ( a OR b ) nor $t0, $t1, $zero Register 0: always read as zero $t0 0000 0000 0000 0000 0011 1101 1100 0000 $t1 0000 0000 0000 0000 0011 1100 0000 0000 $t0 1111 1111 1111 1111 1100 0011 1111 1111 Chapter 2 Instructions: Language of the Computer 77 Chapter 2 Instructions: Language of the Computer 78 2

Conditional Operations Branch to a labeled instruction if a condition is true Otherwise, continue sequentially beq rs, rt, L1 if (rs == rt) branch to instruction labeled L1; bne rs, rt, L1 if (rs!= rt) branch to instruction labeled L1; j L1 unconditional jump to instruction labeled L1 2.7 Instructions for Making Decisions Compiling If Statements C code: if (i==j) f = g+h; else f = g-h; f, g,h in $s0, $s1, $s2 Compiled MIPS code: bne $s3, $s4, Else add $s0, $s1, $s2 j Exit Else: sub $s0, $s1, $s2 Exit: Assembler calculates addresses Chapter 2 Instructions: Language of the Computer 79 Chapter 2 Instructions: Language of the Computer 80 Compiling Loop Statements C code: while (save[i] == k) i += 1; i in $s3, k in $s5, address of save in $s6 Compiled MIPS code: Loop: sll $t1, $s3, 2 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit addi $s3, $s3, 1 j Loop Exit: Basic Blocks A basic block is a sequence of instructions with No embedded branches (except at end) No branch targets (except at beginning) A compiler identifies basic blocks for optimization An advanced processor can accelerate execution of basic blocks Chapter 2 Instructions: Language of the Computer 81 Chapter 2 Instructions: Language of the Computer 82 More Conditional Operations Set result to 1 if a condition is true Otherwise, set to 0 slt rd, rs, rt if (rs < rt) rd = 1; else rd = 0; slti rt, rs, constant if (rs < constant) rt = 1; else rt = 0; Use in combination with beq, bne slt $t0, $s1, $s2 # if ($s1 < $s2) bne $t0, $zero, L # branch to L Branch Instruction Design Why not blt, bge, etc? Hardware for <,, slower than =, Combining with branch involves more work per instruction, requiring a slower clock All instructions penalized! beq and bne are the common case This is a good design compromise Chapter 2 Instructions: Language of the Computer 83 Chapter 2 Instructions: Language of the Computer 84 3

Signed vs. Unsigned Signed comparison: slt, slti Unsigned comparison: sltu, sltui Example $s0 = 1111 1111 1111 1111 1111 1111 1111 1111 $s1 = 0000 0000 0000 0000 0000 0000 0000 0001 slt $t0, $s0, $s1 # signed 1 < +1 $t0 = 1 sltu $t0, $s0, $s1 # unsigned +4,294,967,295 > +1 $t0 = 0 Procedure Calling Procedure (function) performs a specific task and returns results to caller. Chapter 2 Instructions: Language of the Computer 85 Chapter 2 Instructions: Language of the Computer 86 Procedure Calling Calling program Place parameters in registers $a0 - $a3 Transfer control to procedure Called procedure Acquire storage for procedure, save values of required register(s) on stack $sp Perform procedure s operations, restore the values of registers that it used Place result in register for caller $v0 - $v1 Return to place of call by returning to instruction whose address is saved in $ra Register Usage $a0 $a3: arguments (reg s 4 7) $v0, $v1: result values (reg s 2 and 3) $t0 $t9: temporaries Can be overwritten by callee $s0 $s7: saved Must be saved/restored by callee $gp: global pointer for static data (reg 28) $sp: stack pointer for dynamic data (reg 29) $fp: frame pointer (reg 30) $ra: return address (reg 31) Chapter 2 Instructions: Language of the Computer 87 Chapter 2 Instructions: Language of the Computer 88 Procedure Call Instructions Procedure call: jump and link jal ProcedureLabel Address of following instruction put in $ra Jumps to target address Procedure return: jump register jr $ra Copies $ra to program counter Can also be used for computed jumps e.g., for case/switch statements Leaf Procedure Example C code: int leaf_example (int g, h, i, j) { int f; f = (g + h) - (i + j); return f; } Arguments g,, j in $a0,, $a3 f in $s0 (hence, need to save $s0 on stack) Result in $v0 Chapter 2 Instructions: Language of the Computer 89 Chapter 2 Instructions: Language of the Computer 90 4

Leaf Procedure Example (2) MIPS code: leaf_example: addi $sp, $sp, -4 sw $s0, 0($sp) add $t0, $a0, $a1 add $t1, $a2, $a3 sub $s0, $t0, $t1 add $v0, $s0, $zero lw $s0, 0($sp) addi $sp, $sp, 4 jr $ra Save $s0 on stack Procedure body Result Restore $s0 Return Leaf Procedure Example (3) MIPS code for calling function: main: jal leaf_example Chapter 2 Instructions: Language of the Computer 91 Chapter 2 Instructions: Language of the Computer 92 Non-Leaf Procedures Procedures that call other procedures For nested call, caller needs to save on the stack: Its return address Any arguments and temporaries needed after the call Restore from the stack after the call Non-Leaf Procedure Example (2) C code: int fact (int n) { if (n < 1) return 1; else return n * fact(n - 1); } Argument n in $a0 Result in $v0 Chapter 2 Instructions: Language of the Computer 93 Chapter 2 Instructions: Language of the Computer 94 Non-Leaf Procedure Example (3) MIPS code: fact: addi $sp, $sp, -8 # adjust stack for 2 items sw $ra, 4($sp) # save return address sw $a0, 0($sp) # save argument slti $t0, $a0, 1 # test for n < 1 beq $t0, $zero, L1 addi $v0, $zero, 1 # if so, result is 1 addi $sp, $sp, 8 # pop 2 items from stack jr $ra # and return L1: addi $a0, $a0, -1 # else decrement n jal fact # recursive call lw $a0, 0($sp) # restore original n lw $ra, 4($sp) # and return address addi $sp, $sp, 8 # pop 2 items from stack mul $v0, $a0, $v0 # multiply to get result jr $ra # and return Non-Leaf Procedure Example (4) Chapter 2 Instructions: Language of the Computer 95 Chapter 2 Instructions: Language of the Computer 96 5

Non-Leaf Procedure Example (5) Non-Leaf Procedure Example (6) Chapter 2 Instructions: Language of the Computer 97 Chapter 2 Instructions: Language of the Computer 98 Non-Leaf Procedure Example (7) Non-Leaf Procedure Example (8) 6 Chapter 2 Instructions: Language of the Computer 99 Chapter 2 Instructions: Language of the Computer 100 Local Data on the Stack Local data allocated by callee e.g., C automatic variables Procedure frame (activation record) Used by some compilers to manage stack storage Memory Layout Text: program code Static data: global variables e.g., static variables in C, constant arrays and strings $gp initialized to address allowing ±offsets into this segment Dynamic data: heap E.g., malloc in C, new in Java Stack: automatic storage Chapter 2 Instructions: Language of the Computer 101 Chapter 2 Instructions: Language of the Computer 102 6

Register Summary The following registers are preserved on call $s0 - $s7, $gp, $sp, $fp, and $ra Character Data Byte-encoded character sets ASCII: (7-bit) 128 characters 95 graphic, 33 control Latin-1: (8-bit) 256 characters ASCII, +96 more graphic characters Unicode: 32-bit character set Used in Java, C++ wide characters, Most of the world s alphabets, plus symbols UTF-8, UTF-16: variable-length encodings 2.9 Communicating with People Chapter 2 Instructions: Language of the Computer 103 Chapter 2 Instructions: Language of the Computer 104 ASCII Representation of Characters ASCII Characters American Standard Code for Information Interchange (ASCII). Most computers use 8-bit to represent each character. (Java uses Unicode, which is 16- bit). Signs are combination of characters. How to load a byte? lb, lbu, sb for byte (ASCII) lh, lhu, sh for half-word instruction (Unicode) Chapter 2 Instructions: Language of the Computer 105 Chapter 2 Instructions: Language of the Computer 106 Byte/Halfword Operations Could use bitwise operations MIPS byte/halfword load/store String processing is a common case lb rt, offset(rs) lh rt, offset(rs) Sign extend to 32 bits in rt lbu rt, offset(rs) lhu rt, offset(rs) Zero extend to 32 bits in rt sb rt, offset(rs) sh rt, offset(rs) Store just rightmost byte/halfword String Copy Example C code: Null-terminated string void strcpy (char x[], char y[]) { int i; i = 0; while ((x[i]=y[i])!='\0') i += 1; } Addresses of x, y in $a0, $a1 i in $s0 Chapter 2 Instructions: Language of the Computer 107 Chapter 2 Instructions: Language of the Computer 108 7

String Copy Example MIPS code: strcpy: addi $sp, $sp, -4 # adjust stack for 1 item sw $s0, 0($sp) # save $s0 add $s0, $zero, $zero # i = 0 L1: add $t1, $s0, $a1 # addr of y[i] in $t1 lbu $t2, 0($t1) # $t2 = y[i] add $t3, $s0, $a0 # addr of x[i] in $t3 sb $t2, 0($t3) # x[i] = y[i] beq $t2, $zero, L2 # exit loop if y[i] == 0 addi $s0, $s0, 1 # i = i + 1 j L1 # next iteration of loop L2: lw $s0, 0($sp) # restore saved $s0 addi $sp, $sp, 4 # pop 1 item from stack jr $ra # and return 32-bit Constants Most constants are small 16-bit immediate is sufficient For the occasional 32-bit constant lui rt, constant Copies 16-bit constant to left 16 bits of rt Clears right 16 bits of rt to 0 lui $s0,61 0000 0000 0011 1101 0000 0000 0000 0000 ori $s0,$s0,2304 0000 0000 0011 1101 0000 1001 0000 0000 2.10 MIPS Addressing for 32-Bit Immediates and Addresses Chapter 2 Instructions: Language of the Computer 109 Chapter 2 Instructions: Language of the Computer 110 Branch Addressing Branch instructions specify Opcode, two registers, target address Most branch targets are near branch Forward or backward op rs rt constant or address PC-relative addressing Target address = PC + offset 4 PC already incremented by 4 by this time Chapter 2 Instructions: Language of the Computer 111 Jump Addressing Jump (j and jal) targets could be anywhere in text segment Encode full address in instruction op address 6 bits 26 bits PseudoDirect jump addressing Target address = PC 31 28 : (address 4) 32 bits = 4 bits 28 bits Chapter 2 Instructions: Language of the Computer 112 Target Addressing Example Loop code from earlier example Assume Loop at location 80000 Loop: sll $t1, $s3, 2 80000 0 0 19 9 4 0 add $t1, $t1, $s6 80004 0 9 22 9 0 32 lw $t0, 0($t1) 80008 35 9 8 0 bne $t0, $s5, Exit 80012 5 8 21 2 addi $s3, $s3, 1 80016 8 19 19 1 j Loop 80020 2 20000 Exit: 80024 Chapter 2 Instructions: Language of the Computer 113 Branching Far Away If branch target is too far to encode with 16-bit offset, assembler rewrites the code Example written as beq $s0,$s1, L1 bne $s0,$s1, L2 j L1 L2: Chapter 2 Instructions: Language of the Computer 114 8

Addressing Mode Summary Synchronization (Parallelism) Two processors sharing an area of memory P1 writes, then P2 reads Data race if P1 and P2 don t synchronize Result depends on order of accesses Hardware support required Atomic read/write memory operation No other access to the location allowed between the read and write Could be a single instruction E.g., atomic swap of register memory Or an atomic pair of instructions 2.11 Parallelism and Instructions: Synchronization Chapter 2 Instructions: Language of the Computer 115 Chapter 2 Instructions: Language of the Computer 116 Synchronization in MIPS Load linked: ll rt, offset(rs) Store conditional: sc rt, offset(rs) Succeeds if location not changed since the ll Returns 1 in rt Fails if location is changed Returns 0 in rt Example: atomic swap (to test/set lock variable) try: add $t0,$zero,$s4 ;copy exchange value ll $t1,0($s1) ;load linked sc $t0,0($s1) ;store conditional beq $t0,$zero,try ;branch store fails add $s4,$zero,$t1 ;put load value in $s4 C Sort Example Illustrates use of assembly instructions for a C bubble sort function Swap procedure (leaf) void swap(int v[], int k) { int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; } v in $a0, k in $a1, temp in $t0 2.13 A C Sort Example to Put It All Together Chapter 2 Instructions: Language of the Computer 117 Chapter 2 Instructions: Language of the Computer 118 The Procedure Swap swap: sll $t1, $a1, 2 # $t1 = k * 4 add $t1, $a0, $t1 # $t1 = v+(k*4) # (address of v[k]) lw $t0, 0($t1) # $t0 (temp) = v[k] lw $t2, 4($t1) # $t2 = v[k+1] sw $t2, 0($t1) # v[k] = $t2 (v[k+1]) sw $t0, 4($t1) # v[k+1] = $t0 (temp) jr $ra # return to calling routine Chapter 2 Instructions: Language of the Computer 119 Example.data STR:.asciiz "a1b2c3d4e5f6g7h8i9" # STR[0,1,..,17]=a,1,b,..,9 (8 bits) MAX:.word 0x44556677; # MAX = 0x44556677 (32 bits) SIZE:.byte 33,22,11; # SIZE[0,1,2] = 33,22,11 (8 bits) count:.word 0,1,2; # count[0,1,2] = 0,1,2 (32 bits) #-----------------------------------------------------------------------------------------------------.text main: la $t0, STR # $t0 = address(str) lb $t1, 0($t0) # $t1 = 97 (ascii code for 'a' in decimal) addi $t2, $t1, -4 # $t2 = 93 lb $t3, 3($t0) # $t3 = 50 (ascii code for '2' in decimal) lb $t4, 23($t0) # $t4 = 68 = 44 hex lb $t5, 24($t0) # $t5 = 33 lb $t6, 32($t0) # $t6 = 1 lb $t7, 33($t0) # $t7 = 0 lh $t8, 26($t0) # $t8 = 11 = b hex lw $t9, 36($t0) # $t9 = 2 #----------------------------------------------------------------------------------------------------- jr $ra # return Chapter 2 Instructions: Language of the Computer 120 9

Concluding Remarks Design principles 1. Simplicity favors regularity 2. Smaller is faster 3. Make the common case fast 4. Good design demands good compromises Layers of software/hardware Compiler, assembler, hardware MIPS: typical of RISC ISAs c.f. x86 2.20 Concluding Remarks Acknowledgement The slides are adopted from Computer Organization and Design, 5th Edition by David A. Patterson and John L. Hennessy 2014, published by MK (Elsevier) Chapter 2 Instructions: Language of the Computer 121 Chapter 2 Instructions: Language of the Computer 121 10