CSC 320 Lecture Chapter 2, Fall 2012

CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 1 Chapter 2 Instructions: Language of the Computer Chapter 2 — Instructions: Language of the Computer — 2 Instruction Set � The repertoire of instructions of a computer � Different computers have different instruction sets � But with many aspects in common � Early computers had very simple instruction sets � Simplified implementation � Many modern computers also have simple instruction sets §2.1 Introduction CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 2 Chapter 2 — Instructions: Language of the Computer — 3 The MIPS Instruction Set � Used as the example throughout the book � Stanford MIPS commercialized by MIPS Technologies (www.mips.com) � Large share of embedded core market � Applications in consumer electronics, network/storage equipment, cameras, printers, … � Typical of many modern ISAs � See MIPS Reference Data tear-out card, and Appendixes B and E Chapter 2 — Instructions: Language of the Computer — 4 Arithmetic Operations � Add and subtract, three operands � Two sources and one destination add a, b, c # a gets b + c � All arithmetic operations have this form � Design Principle 1: Simplicity favours regularity � Regularity makes implementation simpler � Simplicity enables higher performance at lower cost §2.2 Operations of the Computer Hardware CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 3 Chapter 2 — Instructions: Language of the Computer — 5 Arithmetic Example � C code: f = (g + h) - (i + j); � Compiled MIPS code: add t0, g, h # temp t0 = g + h add t1, i, j # temp t1 = i + j sub f, t0, t1 # f = t0 - t1 Chapter 2 — Instructions: Language of the Computer — 6 Register Operands � Arithmetic instructions use register operands � MIPS has a 32 × 32-bit register file � Use for frequently accessed data � Numbered 0 to 31 � 32-bit data called a “word” � Assembler names � $t0, $t1, …, $t9 for temporary values � $s0, $s1, …, $s7 for saved variables � Design Principle 2: Smaller is faster � c.f. main memory: millions of locations §2.3 Operands of the Computer Hardware CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 4 Chapter 2 — Instructions: Language of the Computer — 7 Register Operand Example � C code: f = (g + h) - (i + j); � f, …, j in $s0, …, $s4 � Compiled MIPS code: add $t0, $s1, $s2 add $t1, $s3, $s4 sub $s0, $t0, $t1 Chapter 2 — Instructions: Language of the Computer — 8 Memory Operands � Main memory used for composite data � Arrays, structures, dynamic data � To apply arithmetic operations � Load values from memory into registers � Store result from register to memory � Memory is byte addressed � Each address identifies an 8-bit byte � Words are aligned in memory � Address must be a multiple of 4 � MIPS is Big Endian � Most-significant byte at least address of a word � c.f. Little Endian: least-significant byte at least address CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 5 Chapter 2 — Instructions: Language of the Computer — 9 Memory Operand Example 1 � C code: g = h + A[8]; � g in $s1, h in $s2, base address of A in $s3 � Compiled MIPS code: � Index 8 requires offset of 32 � 4 bytes per word lw $t0, 32($s3) # load word add $s1, $s2, $t0 offset base register Chapter 2 — Instructions: Language of the Computer — 10 Memory Operand Example 2 � C code: A[12] = h + A[8]; � h in $s2, base address of A in $s3 � Compiled MIPS code: � Index 8 requires offset of 32 lw $t0, 32($s3) # load word add $t0, $s2, $t0 sw $t0, 48($s3) # store word CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 6 Chapter 2 — Instructions: Language of the Computer — 11 Registers vs. Memory � Registers are faster to access than memory � Operating on memory data requires loads and stores � More instructions to be executed � Compiler must use registers for variables as much as possible � Only spill to memory for less frequently used variables � Register optimization is important! Chapter 2 — Instructions: Language of the Computer — 12 Immediate Operands � Constant data specified in an instruction addi $s3, $s3, 4 � No subtract immediate instruction � Just use a negative constant addi $s2, $s1, -1 � Design Principle 3: Make the common case fast � Small constants are common � Immediate operand avoids a load instruction CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 7 Chapter 2 — Instructions: Language of the Computer — 13 The Constant Zero � MIPS register 0 ($zero) is the constant 0 � Cannot be overwritten � Useful for common operations � E.g., move between registers add $t2, $s1, $zero Chapter 2 — Instructions: Language of the Computer — 14 Unsigned Binary Integers � Given an n-bit number 0 0 1 1 n 2 n 2 n 1 n 1 x 2 x 2 2 x 2 x x + + + + = − − − − L � Range: 0 to +2n – 1 � Example � 0000 0000 0000 0000 0000 0000 0000 10112 = 0 + … + 1×23 + 0×22 +1×21 +1×20 = 0 + … + 8 + 0 + 2 + 1 = 1110 � Using 32 bits � 0 to +4,294,967,295 §2.4 Signed and Unsigned Numbers CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 8 Chapter 2 — Instructions: Language of the Computer — 15 2s-Complement Signed Integers � Given an n-bit number 0 0 1 1 n 2 n 2 n 1 n 1 x 2 x 2 2 x 2 x x + + + + = − − − − − L � Range: –2n – 1 to +2n – 1 – 1 � Example � 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 � Using 32 bits � –2,147,483,648 to +2,147,483,647 Chapter 2 — Instructions: Language of the Computer — 16 2s-Complement Signed Integers � Bit 31 is sign bit � 1 for negative numbers � 0 for non-negative numbers � –(–2n – 1) can’t be represented � Non-negative numbers have the same unsigned and 2s-complement representation � Some specific numbers � 0: 0000 0000 … 0000 � –1: 1111 1111 … 1111 � Most-negative: 1000 0000 … 0000 � Most-positive: 0111 1111 … 1111 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 9 Chapter 2 — Instructions: Language of the Computer — 17 Signed Negation � Complement and add 1 � Complement means 1 → 0, 0 → 1 x 1 x 1 1111...111 x x 2 = − + = − = + � Example: negate +2 � +2 = 0000 0000 … 00102 � –2 = 1111 1111 … 11012 + 1 = 1111 1111 … 11102 Chapter 2 — Instructions: Language of the Computer — 18 Sign Extension � Representing a number using more bits � Preserve the numeric value � In MIPS instruction set � addi: extend immediate value � lb, lh: extend loaded byte/halfword � beq, bne: extend the displacement � Replicate the sign bit to the left � c.f. unsigned values: extend with 0s � Examples: 8-bit to 16-bit � +2: 0000 0010 => 0000 0000 0000 0010 � –2: 1111 1110 => 1111 1111 1111 1110 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 10 Chapter 2 — Instructions: Language of the Computer — 19 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 Chapter 2 — Instructions: Language of the Computer — 20 MIPS R-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) op rs rt rd shamt funct 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 11 Chapter 2 — Instructions: Language of the Computer — 21 R-format Example add $t0, $s1, $s2 special $s1 $s2 $t0 0 add 0 17 18 8 0 32 000000 10001 10010 01000 00000 100000 000000100011001001000000001000002 = 0232402016 op rs rt rd shamt funct 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits Chapter 2 — Instructions: Language of the Computer — 22 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 12 Chapter 2 — Instructions: Language of the Computer — 23 MIPS I-format Instructions � Immediate arithmetic and load/store instructions � rt: destination or source register number � Constant: –215 to +215 – 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 6 bits 5 bits 5 bits 16 bits Chapter 2 — Instructions: Language of the Computer — 24 Stored Program Computers � 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 The BIG Picture CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 13 Chapter 2 — Instructions: Language of the Computer — 25 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 — 26 Shift Operations � shamt: how many positions to shift � Shift left logical � Shift left and fill with 0 bits � sll by i bits multiplies by 2i � Shift right logical � Shift right and fill with 0 bits � srl by i bits divides by 2i (unsigned only) op rs rt rd shamt funct 6 bits 6 bits 5 bits 5 bits 5 bits 5 bits CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 14 Chapter 2 — Instructions: Language of the Computer — 27 AND Operations � Useful to mask bits in a word � Select some bits, clear others to 0 and $t0, $t1, $t2 0000 0000 0000 0000 0000 1101 1100 0000 0000 0000 0000 0000 0011 1100 0000 0000 $t2 $t1 0000 0000 0000 0000 0000 1100 0000 0000 $t0 Chapter 2 — Instructions: Language of the Computer — 28 OR Operations � Useful to include bits in a word � Set some bits to 1, leave others unchanged or $t0, $t1, $t2 0000 0000 0000 0000 0000 1101 1100 0000 0000 0000 0000 0000 0011 1100 0000 0000 $t2 $t1 0000 0000 0000 0000 0011 1101 1100 0000 $t0 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 15 Chapter 2 — Instructions: Language of the Computer — 29 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 0000 0000 0000 0000 0011 1100 0000 0000 $t1 1111 1111 1111 1111 1100 0011 1111 1111 $t0 Register 0: always read as zero Chapter 2 — Instructions: Language of the Computer — 30 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 16 Chapter 2 — Instructions: Language of the Computer — 31 Compiling If Statements � C code: if (i==j) f = g+h; else f = g-h; � f, g, … in $s0, $s1, … � 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 — 32 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: … CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 17 Chapter 2 — Instructions: Language of the Computer — 33 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 — 34 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 18 Chapter 2 — Instructions: Language of the Computer — 35 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 — 36 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 19 Chapter 2 — Instructions: Language of the Computer — 37 Procedure Calling � Steps required 1. Place parameters in registers 2. Transfer control to procedure 3. Acquire storage for procedure 4. Perform procedure’s operations 5. Place result in register for caller 6. Return to place of call §2.8 Supporting Procedures in Computer Hardware Chapter 2 — Instructions: Language of the Computer — 38 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 (reg 29) � $fp: frame pointer (reg 30) � $ra: return address (reg 31) CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 20 Chapter 2 — Instructions: Language of the Computer — 39 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 Chapter 2 — Instructions: Language of the Computer — 40 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 21 Chapter 2 — Instructions: Language of the Computer — 41 Leaf Procedure Example � 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 Restore $s0 Result Return Chapter 2 — Instructions: Language of the Computer — 42 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 22 Chapter 2 — Instructions: Language of the Computer — 43 Non-Leaf Procedure Example � C code: int fact (int n) { if (n < 1) return f; else return n * fact(n - 1); } � Argument n in $a0 � Result in $v0 Chapter 2 — Instructions: Language of the Computer — 44 Non-Leaf Procedure Example � 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 23 Chapter 2 — Instructions: Language of the Computer — 45 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 Chapter 2 — Instructions: Language of the Computer — 46 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 24 Chapter 2 — Instructions: Language of the Computer — 47 Character Data � Byte-encoded character sets � ASCII: 128 characters � 95 graphic, 33 control � Latin-1: 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 — 48 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 25 Chapter 2 — Instructions: Language of the Computer — 49 String Copy Example � C code (naïve): � 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 — 50 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 26 Chapter 2 — Instructions: Language of the Computer — 51 0000 0000 0111 1101 0000 0000 0000 0000 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 lhi $s0, 61 0000 0000 0111 1101 0000 1001 0000 0000 ori $s0, $s0, 2304 §2.10 MIPS Addressing for 32-Bit Immediates and Addresses Chapter 2 — Instructions: Language of the Computer — 52 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 6 bits 5 bits 5 bits 16 bits � PC-relative addressing � Target address = PC + offset × 4 � PC already incremented by 4 by this time CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 27 Chapter 2 — Instructions: Language of the Computer — 53 Jump Addressing � Jump (j and jal) targets could be anywhere in text segment � Encode full address in instruction op address 6 bits 26 bits � (Pseudo)Direct jump addressing � Target address = PC31…28 : (address × 4) Chapter 2 — Instructions: Language of the Computer — 54 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 28 Chapter 2 — Instructions: Language of the Computer — 55 Branching Far Away � If branch target is too far to encode with 16-bit offset, assembler rewrites the code � Example beq $s0,$s1, L1 ↓ bne $s0,$s1, L2 j L1 L2: … Chapter 2 — Instructions: Language of the Computer — 56 Addressing Mode Summary CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 29 Chapter 2 — Instructions: Language of the Computer — 57 Synchronization � Two processors sharing an area of memory � P1 writes, then P2 reads � Data race if P1 and P2 don’t synchronize � Result depends of 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 — 58 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 30 Chapter 2 — Instructions: Language of the Computer — 59 Translation and Startup Many compilers produce object modules directly Static linking §2.12 Translating and Starting a Program Chapter 2 — Instructions: Language of the Computer — 60 Assembler Pseudoinstructions � Most assembler instructions represent machine instructions one-to-one � Pseudoinstructions: figments of the assembler’s imagination move $t0, $t1 → add $t0, $zero, $t1 blt $t0, $t1, L → slt $at, $t0, $t1 bne $at, $zero, L � $at (register 1): assembler temporary CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 31 Chapter 2 — Instructions: Language of the Computer — 61 Producing an Object Module � Assembler (or compiler) translates program into machine instructions � Provides information for building a complete program from the pieces � Header: described contents of object module � Text segment: translated instructions � Static data segment: data allocated for the life of the program � Relocation info: for contents that depend on absolute location of loaded program � Symbol table: global definitions and external refs � Debug info: for associating with source code Chapter 2 — Instructions: Language of the Computer — 62 Linking Object Modules � Produces an executable image 1. Merges segments 2. Resolve labels (determine their addresses) 3. Patch location-dependent and external refs � Could leave location dependencies for fixing by a relocating loader � But with virtual memory, no need to do this � Program can be loaded into absolute location in virtual memory space CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 32 Chapter 2 — Instructions: Language of the Computer — 63 Loading a Program � Load from image file on disk into memory 1. Read header to determine segment sizes 2. Create virtual address space 3. Copy text and initialized data into memory � Or set page table entries so they can be faulted in 4. Set up arguments on stack 5. Initialize registers (including $sp, $fp, $gp) 6. Jump to startup routine � Copies arguments to $a0, … and calls main � When main returns, do exit syscall Chapter 2 — Instructions: Language of the Computer — 64 Dynamic Linking � Only link/load library procedure when it is called � Requires procedure code to be relocatable � Avoids image bloat caused by static linking of all (transitively) referenced libraries � Automatically picks up new library versions CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 33 Chapter 2 — Instructions: Language of the Computer — 65 Lazy Linkage Indirection table Stub: Loads routine ID, Jump to linker/loader Linker/loader code Dynamically mapped code Chapter 2 — Instructions: Language of the Computer — 66 Starting Java Applications Simple portable instruction set for the JVM Interprets bytecodes Compiles bytecodes of “hot” methods into native code for host machine CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 34 Chapter 2 — Instructions: Language of the Computer — 67 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 — 68 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 CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 35 Chapter 2 — Instructions: Language of the Computer — 69 The Sort Procedure in C � Non-leaf (calls swap) void sort (int v[], int n) { int i, j; for (i = 0; i < n; i += 1) { for (j = i – 1; j >= 0 && v[j] > v[j + 1]; j -= 1) { swap(v,j); } } } � v in $a0, k in $a1, i in $s0, j in $s1 Chapter 2 — Instructions: Language of the Computer — 70 The Procedure Body move $s2, $a0 # save $a0 into $s2 move $s3, $a1 # save $a1 into $s3 move $s0, $zero # i = 0 for1tst: slt $t0, $s0, $s3 # $t0 = 0 if $s0 ≥ $s3 (i ≥ n) beq $t0, $zero, exit1 # go to exit1 if $s0 ≥ $s3 (i ≥ n) addi $s1, $s0, –1 # j = i – 1 for2tst: slti $t0, $s1, 0 # $t0 = 1 if $s1 < 0 (j < 0) bne $t0, $zero, exit2 # go to exit2 if $s1 < 0 (j < 0) sll $t1, $s1, 2 # $t1 = j * 4 add $t2, $s2, $t1 # $t2 = v + (j * 4) lw $t3, 0($t2) # $t3 = v[j] lw $t4, 4($t2) # $t4 = v[j + 1] slt $t0, $t4, $t3 # $t0 = 0 if $t4 ≥ $t3 beq $t0, $zero, exit2 # go to exit2 if $t4 ≥ $t3 move $a0, $s2 # 1st param of swap is v (old $a0) move $a1, $s1 # 2nd param of swap is j jal swap # call swap procedure addi $s1, $s1, –1 # j –= 1 j for2tst # jump to test of inner loop exit2: addi $s0, $s0, 1 # i += 1 j for1tst # jump to test of outer loop Pass params & call Move params Inner loop Outer loop Inner loop Outer loop CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 36 Chapter 2 — Instructions: Language of the Computer — 71 sort: addi $sp,$sp, –20 # make room on stack for 5 registers sw $ra, 16($sp) # save $ra on stack sw $s3,12($sp) # save $s3 on stack sw $s2, 8($sp) # save $s2 on stack sw $s1, 4($sp) # save $s1 on stack sw $s0, 0($sp) # save $s0 on stack … # procedure body … exit1: lw $s0, 0($sp) # restore $s0 from stack lw $s1, 4($sp) # restore $s1 from stack lw $s2, 8($sp) # restore $s2 from stack lw $s3,12($sp) # restore $s3 from stack lw $ra,16($sp) # restore $ra from stack addi $sp,$sp, 20 # restore stack pointer jr $ra # return to calling routine The Full Procedure Chapter 2 — Instructions: Language of the Computer — 72 Effect of Compiler Optimization 0 0.5 1 1.5 2 2.5 3 none O1 O2 O3 Relative Performance 0 20000 40000 60000 80000 100000 120000 140000 160000 180000 none O1 O2 O3 Clock Cycles 0 20000 40000 60000 80000 100000 120000 140000 none O1 O2 O3 Instruction count 0 0.5 1 1.5 2 none O1 O2 O3 CPI Compiled with gcc for Pentium 4 under Linux CSC320: Computer Organization 27 February 2012 Chapter 2 — Instructions: Language of the Computer 37 Chapter 2 — Instructions: Language of the Computer — 73 Effect of Language and Algorithm 0 0.5 1 1.5 2 2.5 3 C/none C/O1 C/O2 C/O3 Java/int Java/JIT Bubblesort Relative Performance 0 0.5 1 1.5 2 2.5 C/none C/O1 C/O2 C/O3 Java/int Java/JIT Quicksort Relative Performance 0 500 1000 1500 2000 2500 3000 C/none C/O1 C/O2 C/O3 Java/int Java/JIT Quicksort vs. Bubblesort Speedup Chapter 2 — Instructions: Language of the Computer — 74 Lessons Learnt � Instruction count and CPI are not good performance indicators in isolation � Compiler optimizations are sensitive to the algorithm � Java/JIT compiled code is significantly faster than JVM interpreted � Comparable to optimized C in some case