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Chapter 2 1 Chapter 2 2.1 Introduction • • Language of the Machine We’ll be working with the MIPS instruction set architecture – similar to other architectures developed since the 1980's – Almost 100 million MIPS processors manufactured in 2002 1400 – used by NEC, Nintendo, Cisco, Silicon Graphics, Sony, … Other 1300 1200 1100 1000 SPARC Hitachi SH PowerPC Motorola 68K MIPS 900 IA-32 800 ARM 700 600 500 400 300 200 100 0 1998 1999 2000 2001 2002 2 2.2 Operations of the Computer Hardware • • All instructions have 3 operands Operand order is fixed (destination first) Example: C code: a = b + c MIPS ‘code’: add a, b, c #the sum of b and c is placed in a. (we’ll talk about registers in a bit) “The natural number of operands for an operation like addition is three…requiring every instruction to have exactly three operands, no more and no less, conforms to the philosophy of keeping the hardware simple” 3 MIPS arithmetic • • Design Principle: simplicity favors regularity. Of course this complicates some things... C code: a = b + c + d + e; MIPS code: add a, b, c add a, a, d add a, a, e Example: C code: MIPS code: a = b + c; d = a - e; add a, b, c sub d, a, e 4 Example: C code: f = (g + h) – (i + j); MIPS code: add t0, g, h add t1, i, j sub f, t0, t1 5 2.3 Operands of the Computer Hardware • • Operands of arithmetic instructions must be registers, only 32 registers provided Each register contains 32 bits • Design Principle: smaller is faster. Why? Example: Compiling a C Assignment using registers f = (g + h) – (i + j) variables f, g, h, I and j are assigned to registers: $s0, $s1, $s2, $s3, and $s4 respectively MIPS Code: add $t0, $s1, $s2 add $t1, $s3, $s4 sub $s0, $t0, $t1 6 Memory Operands Data Transfer instructions • Arithmetic instructions operands must be registers, — only 32 registers provided • Compiler associates variables with registers • What about programs with lots of variables Control Input Memory Datapath Processor Output I/O 7 Memory Organization • • • Viewed as a large, single-dimension array, with an address. A memory address is an index into the array "Byte addressing" means that the index points to a byte of memory. 0 1 2 3 4 5 6 ... 8 bits of data 8 bits of data 8 bits of data 8 bits of data 8 bits of data 8 bits of data 8 bits of data 8 Example: g = h + A[8]; Where g $s1 h $s2 base address of A $s3 MIPS Code: lw $t0, 8($s3) # error add $s1, $s2, $t0 9 Hardware Software Interface Alignment restriction • Bytes are nice, but most data items use larger "words" • For MIPS, a word is 32 bits or 4 bytes. 0 4 8 12 ... • • • 32 bits of data 32 bits of data 32 bits of data Registers hold 32 bits of data 32 bits of data 232 bytes with byte addresses from 0 to 232-1 230 words with byte addresses 0, 4, 8, ... 232-4 Words are aligned i.e., what are the least 2 significant bits of a word address? 10 Instructions • Load and store instructions Example: C code: A[12] = h + A[8]; where: h $s2 base address of A $s3 MIPS code: • • • lw $t0, 32($s3) add $t0, $s2, $t0 sw $t0, 48($s3) Can refer to registers by name (e.g., $s2, $t2) instead of number Store word has destination last Remember arithmetic operands are registers, not memory! Can’t write: add 48($s3), $s2, 32($s3) 11 Constant or Immediate Operands Constant in memory: lw $t0, AddrConstant4($s1) add $s3, $s3, $t0 #$t0 = constant 4 #$s3 = $s3 + $t0 Constant operand (add immediate): addi $s3, $s3, 4 #$s3 = $s3 + 4 12 So far we’ve learned: • MIPS — loading words but addressing bytes — arithmetic on registers only • Instruction Meaning add $s1, $s2, $s3 sub $s1, $s2, $s3 lw $s1, 100($s2) sw $s1, 100($s2) $s1 = $s2 + $s3 $s1 = $s2 – $s3 $s1 = Memory[$s2+100] Memory[$s2+100] = $s1 13 2.4 Representing Instructions in the Computer • Instructions, like registers and words of data, are also 32 bits long Example: add $t0, $s1, $s2 – registers have numbers, $t0=8, $s1=17, $s2=18 Instruction Format (MIPS Fields): 000000 10001 op 6 bits op: rs: rt: rd: shamt: funct: rs 5 bits 10010 01000 00000 100000 rt 5 bits rd 5 bits shamt 5 bits funct 6 bits opcode. first register source second register source register destination shift amount function code 14 R-format and I-format • • • Consider the load-word and store-word instructions, – What would the regularity principle have us do? – New principle: Good design demands a compromise Introduce a new type of instruction format – I-type for data transfer instructions – other format was R-type for register Example: lw $t0, 32($s2) 35 18 8 op rs rt 6 bits • 5 bits 5 bits 32 constant or address 16 bits Where's the compromise? 15 Translating MIPS Assembly into Machine Language Example: A[300] = h + A[300]; $s2 $t1 is compiled into: lw $t0, 1200($t1) add $t0, $s2, $t0 sw $t0, 1200($t1) ---------t0 8 t1 9 s2 18 100011 01001 01000 000000 10010 01000 101011 01001 01000 0000 0100 1011 0000 01000 00000 100000 0000 0100 1011 0000 16 So far we’ve learned: 17 Stored Program Concept • • Instructions are bits Programs are stored in memory — to be read or written just like data memory for data, programs, compilers, editors, etc. • Fetch & Execute Cycle – Instructions are fetched and put into a special register – Bits in the register "control" the subsequent actions – Fetch the “next” instruction and continue 18 2.5 Logical Operations Example: sll $t2, $s0, 4 # reg $t2 = reg $s0 << 4 bits $s0 contained: 0000 0000 0000 0000 0000 0000 0000 1001 After executions: 0000 0000 0000 0000 0000 0000 1001 0000 Op Rs Rt Rd shamt funct 0 0 16 $s0 10 $t2 4 0 19 Logical opeartions: and or nor $t1: $t2: $t1 & $t2 $t1 | $t2 0000 0000 0000 0000 $t1: $t3: ~($t1 | $t3) 0000 0000 0000 0000 0011 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 1111 1100 0011 1111 1111 and $t0, $t1, $t2 or $t0, $t1, $t2 nor $t0, $t1, $t3 • • • 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0000 0000 0011 1100 1101 1100 1101 0000 0000 0000 0000 0000 0000 0000 0000 # $t0 = $t1 & $t2 # $t0 = $t1 | $t2 # $t0 = ~($t1 | $t3) andi: and immediate ori : or immediate No immediate version for NOR (its main use is to invert the bits of a single operand. 20 So far we’ve learned: 21 2.6 Instruction for Making Decisions • Decision making instructions – alter the control flow, – i.e., change the "next" instruction to be executed • MIPS conditional branch instructions: beq $t0, $t1, Label bne $t0, $t1, Label • Example: if (i==j) h = i + j; i$s0 i$s1 h$s3 bne $s0, $s1, Label add $s3, $s0, $s1 Label: .... 22 If-else • MIPS unconditional branch instructions: j label Example: f ≡s0 g ≡s1 h ≡s2 i ≡s3 and j ≡s4 if (i==j) f=g+h; else f=g-h; • bne $s3, $s4, Else add $s0, $s1, $s2 j Exit Else: sub $s0, $s1, $s2 Exit: ... Can you build a simple for loop? 23 loops Example: Compiling a while loop in C: While (save[i] == k) i += 1; Assume: i ≡s3 k ≡s5 base-of A[]≡s6 Loop: sll add lw bne add j $t1, $s3, 2 $t1, $t1, $s6 $$t0, 0($t1) $t0, $$5, Exit $s3, $s3, 1 loop Exit: 24 So far: • • Instruction Meaning add $s1,$s2,$s3 sub $s1,$s2,$s3 lw $s1,100($s2) sw $s1,100($s2) bne $s4,$s5,L beq $s4,$s5,L j Label $s1 = $s2 + $s3 $s1 = $s2 – $s3 $s1 = Memory[$s2+100] Memory[$s2+100] = $s1 Next instr. is at Label if $s4 ≠ $s5 Next instr. is at Label if $s4 = $s5 Next instr. is at Label Formats: R op rs rt rd I op rs rt 16 bit address J op shamt funct 26 bit address 25 Continue set on less than instruction: slt $t0, $s3, $s4 slti $t0, $s2, 10 Means: if($s3<$s4) $t0=1; else $t0=0; 26 2.7 Supporting Procedures in Computer Hardware • • • • • $a0-$a3: $v0-Sv1: $ra: jal: jr $ra: four arguments two value registers in which to return values one return address register jump and save return address in $ra. jump to the address stored in register $ra. Using More Registers • Spill registers to memory • Stack • $sp 27 Example: Leaf procedure int leaf_example(int g, int h, int i, int j) { int f; f=(g+h)-(i+j); return f; } leaf_example: addi $sp, $sp, -12 sw $t1, 8($sp) sw $t0, 4($sp) sw $s0, 0($sp) add add sub add $t0, $t1, $s0, $v0, $a0, $a2, $t0, $s0, $a1 $a3 $t1 $zero lw $s0, 0($sp) lw $t0, 4($sp) lw $t1, 8($sp) addi $sp, $sp, 12 jr $ra 28 Nested Procedures int fact (int n) { if(n<1) return (1); else return (n*fact(n-1)); } fact: addi $sp, $sp, -8 sw $ra, 4($sp) sw $a0, 0($sp) slti $t0, $a0, 1 beq $t0, $zero, L1 addi $v0, $zero, 1 addi $sp, $sp, 8 jr $ra L1: addi $a0, $a0, -1 jal fact lw $a0, 0($sp) lw $ra, 4($sp) addi $sp, $sp, 8 mul $v0, $a0, $v0 jr $ra 29 Allocating Space for New Data on the Stack • • • Stack is also used to store Local variables that do not fit in registers (arrays or structures) Procedure frame (activation record) Frame Pointer 30 Allocating Space for New Data on the Heap • Need space for static variables and dynamic data structures (heap). 31 Policy of Use Conventions Name Register number $zero 0 $v0-$v1 2-3 $a0-$a3 4-7 $t0-$t7 8-15 $s0-$s7 16-23 $t8-$t9 24-25 $gp 28 $sp 29 $fp 30 $ra 31 Usage the constant value 0 values for results and expression evaluation arguments temporaries saved more temporaries global pointer stack pointer frame pointer return address Register 1 ($at) reserved for assembler, 26-27 for operating system 32 2.8 Communicating with People • For text process, MIPS provides instructions to move bytes (8 bits ASCII): lb $t0, 0($sp) Load a byte from memory, placing it in the rightmost 8 bits of a register. sb $t0, 0($gp) Store byte (rightmost 8 bits of a register) in to memory. • Java uses Unicode for characters: 16 bits to represent characters: lh $t0, 0($sp) #Read halfword (16 bits) from source sh $t0, 0($gp) #Write halfword (16 bits) to destination 33 Example: void strcpy(char x[], char y[]) { int i; i=0; while((x[i]=y[i])!=‘\0’) i+=1; } strcpy: addi $sp, $sp, -4 sw add L1: add lb add sb beq addi j L2: lw addi jr $s0, $s0, $t1, $t2, $t3, $t2, $t2, $s0, L1 $s0, $sp, $ra 0($sp) $zero, $zero $s0, $a1 0($t1) $s0, $a0 0($t3) $zero, L2 $s0, 1 0($sp) $sp, 4 34 2.9 MIPS Addressing for 32-Bit Immediates and Address 32-Bit Immediate Operands • load upper immediate lui to set the upper 16 bits of a constant in a register. lui $t0, 255 # $t0 is register 8: 001111 00000 01000 0000 0000 1111 1111 Contents of register $t0 after executing: lui $t0, 255 0000 0000 1111 1111 0000 0000 0000 0000 35 Example: loading a 32-Bit Constant What is the MIPS assembly code to load this 32-bit constant into register $s0 0000 0000 0011 1101 0000 1001 0000 0000 lui $s0, 61 The value of $s0 afterward is: 0000 0000 0011 1101 0000 0000 0000 0000 ori $s0, $s0, 2304 The final value in register $s0 is the desire value: 0000 0000 0011 1101 0000 1001 0000 0000 36 Addressing in Branches and Jumps • jump addressing j • 10000 #go to location 10000 2 10000 6 bits 26 bits Unlike the jump, the conditional branch must specify two operands: bne $s0, $s1, Exit # go to Exit if $s0 ≠ $s1 5 16 17 Exit 6 bits 5 bits 5 bits 16 bits • • No program could be bigger than 216, which is far too small Solution: PC-relative addressing Program counter = PC + Branch address PC points to the next instruction to be executed. 37 Examples: Showing Branch Offset in Machine Language loop: while (save[i] == k) i += 1; sll $t1, $s3, 2 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit addi $s3, $s3, 1 j Loop Exit: 80000 0 0 19 9 2 0 80004 0 9 22 9 0 32 80008 35 9 8 0 80012 5 8 21 2 80016 8 19 19 1 80020 2 80024 ……… 2000 38 Example Branching Far Away Given a branch on register $s0 being equal to $s1, beq $S0, $s1, L1 replace it by a pair of instructions that offers a much greater branching distance. These instructions replace the short-address conditional branch: bne $s0, $s1, L2 j L1 L2: ... 39 MIPS Addressing Modes Summary 1. Immediate addressing op 1. 2. 3. 4. 5. Register addressing Base or displacement addressing Immediate addressing PC-relative addressing Pseudodirect addressing: 26 bits concatenated with the upper bits of the PC. rs rt Immediate 2. Register addressing op rs rt rd ... funct Registers Register 3. Base addressing op rs rt Memory Address + Register Byte Halfword Word 4. PC-relative addressing op rs rt Memory Address PC + Word 5. Pseudodirect addressing op Address PC Memory Word 40 Decoding Machine Language What is the assembly language corresponding to this machine code: 00af8020hex 0000 0000 1010 1111 1000 0000 0010 0000 From fig 2.25, it is an R-format instruction 000000 00101 01111 10000 00000 100000 from fig 2.25, Represent an add instruction add $s0, $a1, $t7 41 Constants • • • Small constants are used quite frequently (50% of operands) e.g., A = A + 5; B = B + 1; C = C - 18; Solutions? Why not? – put 'typical constants' in memory and load them. – create hard-wired registers (like $zero) for constants like one. MIPS Instructions: addi $29, $29, 4 slti $8, $18, 10 andi $29, $29, 6 ori $29, $29, 4 • Design Principle: Make the common case fast. Which format? 42 How about larger constants? • • We'd like to be able to load a 32 bit constant into a register Must use two instructions, new "load upper immediate" instruction lui $t0, 1010101010101010 1010101010101010 • filled with zeros 0000000000000000 Then must get the lower order bits right, i.e., ori $t0, $t0, 1010101010101010 1010101010101010 0000000000000000 0000000000000000 1010101010101010 1010101010101010 1010101010101010 ori 43 Assembly Language vs. Machine Language • • • • Assembly provides convenient symbolic representation – much easier than writing down numbers – e.g., destination first Machine language is the underlying reality – e.g., destination is no longer first Assembly can provide 'pseudoinstructions' – e.g., “move $t0, $t1” exists only in Assembly – would be implemented using “add $t0,$t1,$zero” When considering performance you should count real instructions 44 Other Issues • Discussed in your assembly language programming lab: support for procedures linkers, loaders, memory layout stacks, frames, recursion manipulating strings and pointers interrupts and exceptions system calls and conventions • Some of these we'll talk more about later • We’ll talk about compiler optimizations when we hit chapter 4. 45 Overview of MIPS • • • • • simple instructions all 32 bits wide very structured, no unnecessary baggage only three instruction formats R op rs rt rd I op rs rt 16 bit address J op shamt funct 26 bit address rely on compiler to achieve performance — what are the compiler's goals? help compiler where we can 46 Addresses in Branches and Jumps • • • Instructions: bne $t4,$t5,Label $t5 beq $t4,$t5,Label $t5 j Label op I Formats: op J rs Next instruction is at Label if $t4 ° Next instruction is at Label if $t4 = Next instruction is at Label rt 16 bit address 26 bit address Addresses are not 32 bits — How do we handle this with load and store instructions? 47 Addresses in Branches • • Instructions: bne $t4,$t5,Label beq $t4,$t5,Label Formats: I • • Next instruction is at Label if $t4≠$t5 Next instruction is at Label if $t4=$t5 op rs rt 16 bit address Could specify a register (like lw and sw) and add it to address – use Instruction Address Register (PC = program counter) – most branches are local (principle of locality) Jump instructions just use high order bits of PC – address boundaries of 256 MB 48 To summarize: MIPS operands Name 32 registers Example Comments $s0-$s7, $t0-$t9, $zero, Fast locations for data. In MIPS, data must be in registers to perform $a0-$a3, $v0-$v1, $gp, arithmetic. MIPS register $zero always equals 0. Register $at is $fp, $sp, $ra, $at reserved for the assembler to handle large constants. Memory[0], 2 30 Accessed only by data transfer instructions. MIPS uses byte addresses, so memory Memory[4], ..., words and spilled registers, such as those saved on procedure calls. add MIPS assembly language Example Meaning add $s1, $s2, $s3 $s1 = $s2 + $s3 Three operands; data in registers subtract sub $s1, $s2, $s3 $s1 = $s2 - $s3 Three operands; data in registers $s1 = $s2 + 100 $s1 = Memory[$s2 + 100] Memory[$s2 + 100] = $s1 $s1 = Memory[$s2 + 100] Memory[$s2 + 100] = $s1 Used to add constants Category Arithmetic sequential words differ by 4. Memory holds data structures, such as arrays, Memory[4294967292] Instruction addi $s1, $s2, 100 lw $s1, 100($s2) sw $s1, 100($s2) store word lb $s1, 100($s2) load byte sb $s1, 100($s2) store byte load upper immediate lui $s1, 100 add immediate load word Data transfer Conditional branch Unconditional jump $s1 = 100 * 2 16 Comments Word from memory to register Word from register to memory Byte from memory to register Byte from register to memory Loads constant in upper 16 bits branch on equal beq $s1, $s2, 25 if ($s1 == $s2) go to PC + 4 + 100 Equal test; PC-relative branch branch on not equal bne $s1, $s2, 25 if ($s1 != $s2) go to PC + 4 + 100 Not equal test; PC-relative set on less than slt $s1, $s2, $s3 if ($s2 < $s3) $s1 = 1; else $s1 = 0 Compare less than; for beq, bne set less than immediate slti jump j jr jal jump register jump and link $s1, $s2, 100 if ($s2 < 100) $s1 = 1; Compare less than constant else $s1 = 0 2500 $ra 2500 Jump to target address go to 10000 For switch, procedure return go to $ra $ra = PC + 4; go to 10000 For procedure call 49 1. Immediate addressing op rs rt Immediate 2. Register addressing op rs rt rd ... funct Registers Register 3. Base addressing op rs rt Memory Address + Register Byte Halfword Word 4. PC-relative addressing op rs rt Memory Address PC + Word 5. Pseudodirect addressing op Address PC Memory Word 50 Alternative Architectures • Design alternative: – provide more powerful operations – goal is to reduce number of instructions executed – danger is a slower cycle time and/or a higher CPI –“The path toward operation complexity is thus fraught with peril. To avoid these problems, designers have moved toward simpler instructions” • Let’s look (briefly) at IA-32 51 IA - 32 • • • • • • • • • • • 1978: The Intel 8086 is announced (16 bit architecture) 1980: The 8087 floating point coprocessor is added 1982: The 80286 increases address space to 24 bits, +instructions 1985: The 80386 extends to 32 bits, new addressing modes 1989-1995: The 80486, Pentium, Pentium Pro add a few instructions (mostly designed for higher performance) 1997: 57 new “MMX” instructions are added, Pentium II 1999: The Pentium III added another 70 instructions (SSE) 2001: Another 144 instructions (SSE2) 2003: AMD extends the architecture to increase address space to 64 bits, widens all registers to 64 bits and other changes (AMD64) 2004: Intel capitulates and embraces AMD64 (calls it EM64T) and adds more media extensions “This history illustrates the impact of the “golden handcuffs” of compatibility “adding new features as someone might add clothing to a packed bag” “an architecture that is difficult to explain and impossible to love” 52 IA-32 Overview • • Complexity: – Instructions from 1 to 17 bytes long – one operand must act as both a source and destination – one operand can come from memory – complex addressing modes e.g., “base or scaled index with 8 or 32 bit displacement” Saving grace: – the most frequently used instructions are not too difficult to build – compilers avoid the portions of the architecture that are slow “what the 80x86 lacks in style is made up in quantity, making it beautiful from the right perspective” 53 IA-32 Registers and Data Addressing • Registers in the 32-bit subset that originated with 80386 Name Use 31 0 EAX GPR 0 ECX GPR 1 EDX GPR 2 EBX GPR 3 ESP GPR 4 EBP GPR 5 ESI GPR 6 EDI GPR 7 EIP EFLAGS CS Code segment pointer SS Stack segment pointer (top of stack) DS Data segment pointer 0 ES Data segment pointer 1 FS Data segment pointer 2 GS Data segment pointer 3 Instruction pointer (PC) Condition codes 54 IA-32 Register Restrictions • Registers are not “general purpose” – note the restrictions below 55 IA-32 Typical Instructions • Four major types of integer instructions: – Data movement including move, push, pop – Arithmetic and logical (destination register or memory) – Control flow (use of condition codes / flags ) – String instructions, including string move and string compare 56 IA-32 instruction Formats • Typical formats: (notice the different lengths) a. JE EIP + displacement 4 4 8 Condi- Displacement tion JE b. CALL 8 32 CALL Offset c. MOV 6 MOV EBX, [EDI + 45] 1 1 8 d w r/m Postbyte 8 Displacement d. PUSH ESI 5 3 PUSH Reg e. ADD EAX, #6765 4 3 1 32 ADD Reg w f. TEST EDX, #42 7 1 TEST w Immediate 8 32 Postbyte Immediate 57 Summary • • • Instruction complexity is only one variable – lower instruction count vs. higher CPI / lower clock rate Design Principles: – simplicity favors regularity – smaller is faster – good design demands compromise – make the common case fast Instruction set architecture – a very important abstraction indeed! 58