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Computer Architecture Chapter 3 Instructions: Language of the Machine

Computer Architecture Chapter 3 Instructions: Language of the Machine. Lecture Outline. MIPS operations Arithmetic Logical Data transfer Stacks and Registers Alternative Architectures Summary. Instructions:. Language of the Machine

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Computer Architecture Chapter 3 Instructions: Language of the Machine

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  1. Computer ArchitectureChapter3Instructions: Language of the Machine

  2. Lecture Outline • MIPS operations • Arithmetic • Logical • Data transfer • Stacks and Registers • Alternative Architectures • Summary

  3. Instructions: • Language of the Machine • More primitive than higher level languages e.g., no sophisticated control flow • Very restrictive e.g., MIPS Arithmetic Instructions • We’ll be working with the MIPS instruction set architecture • similar to other architectures developed since the 1980's • used by NEC, Nintendo, Silicon Graphics, Sony Design goals: maximize performance and minimize cost, reduce design time

  4. Generic Examples of Instruction Format Widths … Variable: Fixed: Hybrid: …

  5. Instruction Format • • If code size is most important, use variable length instructions • If performance is most important, use fixed length instructions • Recent embedded machines (ARM, MIPS) added optional mode to execute subset of 16-bit wide instructions (Thumb, MIPS16); per procedure decide performance or density • Some architectures actually exploring on-the-fly decompression for more density.

  6. Typical Operations (little change since 1960) Load (from memory) Store (to memory) memory-to-memory move register-to-register move input (from I/O device) output (to I/O device) push, pop (to/from stack) Data Movement integer (binary + decimal) or FP Add, Subtract, Multiply, Divide Arithmetic shift left/right, rotate left/right Shift not, and, or, set, clear Logical unconditional, conditional Control (Jump/Branch) call, return Subroutine Linkage trap, return Interrupt test & set (atomic r-m-w) Synchronization search, translate String parallel subword ops (4 16bit add) Graphics (MMX)

  7. Top 10 80x86 Instructions

  8. Operation Summary Support these simple instructions, since they will dominate the number of instructions executed: load, store, add, subtract, move register-register, and, shift, compare equal, compare not equal, branch, jump, call, return;

  9. Instruction Fetch Instruction Decode Operand Fetch Execute Result Store Next Instruction Instruction Set Architecture: What Must be Specified? • Instruction Format or Encoding • how is it decoded? • Location of operands and result • where other than memory? • how many explicit operands? • how are memory operands located? • which can or cannot be in memory? • Data type and Size • Operations • what are supported • Successor instruction • jumps, conditions, branches • fetch-decode-execute is implicit!

  10. Processor Memory Stored Program Concept • Instructions are bits • Programs are stored in memory • to be read or written just like data • 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 memory for data, programs, compilers, editors, etc.

  11. Comparison: Bytes per instruction? Number of Instructions? Cycles per instruction? Basic ISA Classes Most real machines are hybrids of these: Accumulator (1 register): 1 address add Aacc ¬ acc + mem[A] 1+x address addx A acc ¬ acc + mem[A + x] Stack: 0 address add tos ¬ tos + next General Purpose Register (can be memory/memory): 2 address add A B EA[A] ¬ EA[A] + EA[B] 3 address add A B C EA[A] ¬ EA[B] + EA[C] Load/Store: 3 address add Ra Rb Rc Ra ¬ Rb + Rc load Ra Rb Ra ¬ mem[Rb] store Ra Rb mem[Rb] ¬ Ra

  12. ° 1975-2000 all machines use general purpose registers ° Advantages of registers • registers are faster than memory • registers are easier for a compiler to use e.g., (A*B) – (C*D) – (E*F) can do multiplies in any order - vs. stack • registers can hold variables - memory traffic is reduced, so program is sped up (since registers are faster than memory) code density improves (since register named with fewer bits - than memory location) General Purpose Registers Dominate Unlike programs in high-level language, the operands of arithmetic instructions cannot be any variables; they must be from a limited number of special locations called registers

  13. Register Register (register-memory) (load-store) Load R1,A Add R1,B Store C, R1 Comparing Number of Instructions Code sequence for (C = A + B) for four classes of instruction sets: Stack Accumulator Push A Load A Load R1,A Push B Add B Load R2,B Add Store C Add R3,R1,R2 Pop C Store C,R3

  14. Historical perspective • Accumulator architectures • General-purpose register architectures • Compact code and stack architecture • High-level-language computer architecture • CISC vs. RISC

  15. Data Types Bit: 0, 1 Bit String: sequence of bits of a particular length 4 bits is a nibble 8 bits is a byte 16 bits is a half-word 32 bits is a word 64 bits is a double-word Character: ASCII 7 bit code Decimal: digits 0-9 encoded as 0000b thru 1001b two decimal digits packed per 8 bit byte Integers: 2's Complement Floating Point: Single Precision Double Precision Extended Precision exponent E M x R base How many +/- #'s? Where is decimal pt? How are +/- exponents represented? mantissa

  16. C Language: C=a+b Control Compile Memory 35 1 5 100 35 2 5 101 Assembly Language: lw $s1, 100($5) lw $s2, 101($5) add $s3, $s2, $s1 sw $s3, 102($5) 0 1 2 3 a: (100($s5)) 36 3 5 102 Datapath b: (101($s5)) Register a $s5+100 c: (102($s5)) Adder a $s5+101 b b a+b $s5+102 a+b Assemble Machine Language: I 35 1 5 100 I 35 2 5 101 R 0 1 2 3 I 36 3 5 102

  17. 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

  18. A B C D Sequential Laundry 2 AM 12 6 PM 1 8 7 11 10 9 • Sequential laundry takes 8 hours for 4 loads • If they learned pipelining, how long would laundry take? 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 T a s k O r d e r Time

  19. 30 30 30 30 30 30 30 A B C D Pipelined Laundry: Start work ASAP 2 AM 12 6 PM 1 8 7 11 10 9 • Pipelined laundry takes 3.5 hours for 4 loads! Time T a s k O r d e r

  20. Pipelining • Improve performance by increasing instruction throughput Ideal speedup is number of stages in the pipeline. Do we achieve this?

  21. Im Dm Reg Reg ALU Im Dm Reg Reg ALU Im Dm Reg Reg ALU Im Dm Reg Reg ALU Im Dm Reg Reg ALU Apply pipelining Time (clock cycles) I n s t r. O r d e r Inst 0 Inst 1 Inst 2 Inst 3 Inst 4

  22. MIPS arithmetic • All instructions have 3 operands • Operand order is fixed (destination first) • Example: C code: A = B + C MIPS code: add $s0, $s1, $s2 (associated with variables by compiler)

  23. MIPS arithmetic • Design Principle: simplicity favors regularity. • Of course this complicates some things... C code: A = B + C + D; E = F - A; MIPS code: add $t0, $s1, $s2 add $s0, $t0, $s3 sub $s4, $s5, $s0 • Operands must be registers, only 32 registers provided • Design Principle: smaller is faster.

  24. MIPS arithmetic instructions Instruction Example Meaning Comments add add $1,$2,$3 $1 = $2 + $3 3 operands; exception possible subtract sub $1,$2,$3 $1 = $2 – $3 3 operands; exception possible add immediate addi $1,$2,100 $1 = $2 + 100 + constant; exception possible add unsigned addu $1,$2,$3 $1 = $2 + $3 3 operands; no exceptions subtract unsigned subu $1,$2,$3 $1 = $2 – $3 3 operands; no exceptions add imm. unsign. addiu $1,$2,100 $1 = $2 + 100 + constant; no exceptions multiply mult $2,$3 Hi, Lo = $2 x $3 64-bit signed product multiply unsigned multu$2,$3 Hi, Lo = $2 x $3 64-bit unsigned product divide div $2,$3 Lo = $2 ÷ $3, Lo = quotient, Hi = remainder Hi = $2 mod $3 divide unsigned divu $2,$3 Lo = $2 ÷ $3, Unsigned quotient & remainder Hi = $2 mod $3 Move from Hi mfhi $1 $1 = Hi Used to get copy of Hi Move from Lo mflo $1 $1 = Lo Used to get copy of Lo

  25. MIPS logical instructions Instruction Example Meaning Comment and and $1,$2,$3 $1 = $2 & $3 3 reg. operands; Logical AND or or $1,$2,$3 $1 = $2 | $3 3 reg. operands; Logical OR xor xor $1,$2,$3 $1 = $2 $3 3 reg. operands; Logical XOR nor nor $1,$2,$3 $1 = ~($2 |$3) 3 reg. operands; Logical NOR and immediate andi $1,$2,10 $1 = $2 & 10 Logical AND reg, constant or immediate ori $1,$2,10 $1 = $2 | 10 Logical OR reg, constant xor immediate xori $1, $2,10 $1 = ~$2 &~10 Logical XOR reg, constant shift left logical sll $1,$2,10 $1 = $2 << 10 Shift left by constant shift right logical srl $1,$2,10 $1 = $2 >> 10 Shift right by constant shift right arithm. sra $1,$2,10 $1 = $2 >> 10 Shift right (sign extend) shift left logical sllv $1,$2,$3 $1 = $2 << $3 Shift left by variable shift right logical srlv $1,$2, $3 $1 = $2 >> $3 Shift right by variable shift right arithm. srav $1,$2, $3 $1 = $2 >> $3 Shift right arith. by variable

  26. Control Input Memory Datapath Output Processor I/O Registers vs. Memory • Arithmetic instructions operands must be registers, • only 32 registers provided • Compiler associates variables with registers • What about programs with lots of variables

  27. 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 8 bits of data 1 8 bits of data 2 8 bits of data 3 8 bits of data 4 8 bits of data 5 8 bits of data 6 8 bits of data ...

  28. Memory Organization • Bytes are nice, but most data items use larger "words" • For MIPS, a word is 32 bits or 4 bytes. • 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? 0 32 bits of data Registers hold 32 bits of data 4 32 bits of data 8 32 bits of data 12 32 bits of data ...

  29. Instructions • Load and store instructions • Example: C code: A[8] = h + A[8]; MIPS code: lw $t0, 32($s3) add $t0, $s2, $t0 sw $t0, 32($s3) • Store word has destination last • Remember arithmetic operands are registers, not memory!

  30. MIPS data transfer instructions Instruction Comment SW 500(R4), R3 Store word SH 502(R2), R3 Store half SB 41(R3), R2 Store byte LW R1, 30(R2) Load word LH R1, 40(R3) Load halfword LHU R1, 40(R3) Load halfword unsigned LB R1, 40(R3) Load byte LBU R1, 40(R3) Load byte unsigned LUI R1, 40 Load Upper Immediate (16 bits shifted left by 16) Why need LUI? LUI R5 R5 0000 … 0000

  31. So far we are learned: • MIPS • loading words but addressing bytes • arithmetic on registers only • InstructionMeaningadd $s1, $s2, $s3 $s1 = $s2 + $s3sub $s1, $s2, $s3 $s1 = $s2 - $s3lw $s1, 100($s2) $s1 = Memory[$s2+100] sw $s1, 100($s2) Memory[$s2+100] = $s1

  32. Machine Language • Instructions, like registers and words of data, are also 32 bits long • Example: add $t0, $s1, $s2 • registers have numbers, $t0=9, $s1=17, $s2=18 • Instruction Format:000000 10001 10010 01000 00000 100000 op rs rt rd shamt funct

  33. Machine Language • 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 9 32 op rs rt 16 bit number

  34. Control • Decision making instructions • alter the control flow, • i.e., change the "next" instruction to be executed • MIPS conditional branch instructions: bne $t0, $t1, Label beq $t0, $t1, Label • Example: if (i==j) h = i + j; bne $s0, $s1, Label add $s3, $s0, $s1 Label: ....

  35. Control • MIPS unconditional branch instructions: j label • Example: if (i!=j) beq $s4, $s5, Lab1 h=i+j; add $s3, $s4, $s5 else j Lab2 h=i-j; Lab1: sub $s3, $s4, $s5 Lab2: ...

  36. op rs rt rd shamt funct op rs rt 16 bit address R I J op 26 bit address So far: • InstructionMeaningadd $s1,$s2,$s3 $s1 = $s2 + $s3sub $s1,$s2,$s3 $s1 = $s2 - $s3lw $s1,100($s2) $s1 = Memory[$s2+100] sw $s1,100($s2) Memory[$s2+100] = $s1bne $s4,$s5,Label Next instr. is at Label if $s4 = $s5beq $s4,$s5,Label Next instr. is at Label if $s4 = $s5j Label Next instr. is at Label • Formats:

  37. 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. • Keep the constant inside the instruction itself (I-type format for immediate.). • MIPS Instructions: addi $29, $29, 4 slti $8, $18, 10 andi $29, $29, 6 ori $29, $29, 4 3

  38. filled with zeros 1010101010101010 0000000000000000 1010101010101010 0000000000000000 0000000000000000 1010101010101010 ori 1010101010101010 1010101010101010 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 • Then must get the lower order bits right, i.e.,ori $t0, $t0, 1010101010101010

  39. Addresses in Branches and Jumps • Instructions: bne $t4,$t5,Label Next instruction is at Label if $t4 = $t5 beq $t4,$t5,Label Next instruction is at Label if $t4 = $t5 j Label Next instruction is at Label • Formats: • Addresses are not 32 bits • How do we handle this with load and store instructions? op rs rt 16 bit address I J op 26 bit address

  40. Overview of MIPS • simple instructions all 32 bits wide • very structured, no unnecessary baggage • only three instruction formats • rely on compiler to achieve performance • what are the compiler's goals? • help compiler where we can op rs rt rd shamt funct R I J op rs rt 16 bit address op 26 bit address

  41. MIPS Compare and Branch (Fixup) • Compare and Branch • BEQ rs, rt, offset if R[rs] == R[rt] then PC-relative branch • BNE rs, rt, offset <> • Compare to zero and Branch • BLEZ rs, offset if R[rs] <= 0 then PC-relative branch • BGTZ rs, offset > • BLT < • BGEZ >= • BLTZAL rs, offset if R[rs] < 0 then branch and link (into R 31) • BGEZAL >= • Remaining set of compare and branch take two instructions • Almost all comparisons are against zero

  42. MIPS jump, branch, compare instructions Instruction Example Meaning branch on equal beq $1,$2,100 if ($1 == $2) go to PC+4+100 Equal test; PC relative branch branch on not eq. bne $1,$2,100 if ($1!= $2) go to PC+4+100 Not equal test; PC relative set on less than slt $1,$2,$3 if ($2 < $3) $1=1; else $1=0 Compare less than; 2’s comp. set less than imm. slti $1,$2,100 if ($2 < 100) $1=1; else $1=0 Compare < constant; 2’s comp. set less than uns. sltu $1,$2,$3 if ($2 < $3) $1=1; else $1=0 Compare less than; natural numbers set l. t. imm. uns. sltiu $1,$2,100 if ($2 < 100) $1=1; else $1=0 Compare < constant; natural numbers jump j 10000 go to 10000 Jump to target address jump register jr $31 go to $31 For switch, procedure return jump and link jal 10000 $31 = PC + 4; go to 10000 For procedure call

  43. R1= 0…00 0000 0000 0000 0001 R2= 0…00 0000 0000 0000 0010 R3= 1…11 1111 1111 1111 1111 After executing these instructions: slt r4,r2,r1 ; if (r2 < r1) r4=1; else r4=0 slt r5,r3,r1 ; if (r3 < r1) r5=1; else r5=0 sltu r6,r2,r1 ; if (r2 < r1) r6=1; else r6=0 sltu r7,r3,r1 ; if (r3 < r1) r7=1; else r7=0 What are values of registers r4 - r7? Why? r4 = ; r5 = ; r6 = ; r7 = ; Signed vs. Unsigned Comparison two two two

  44. Conditional Branch Distance • Distance from branch in instructions 2i  Š ±2i-1 & > 2i-2 • 25% of integer branches are > 2 to 4 or -2 to -4 instructions

  45. Addressing Objects: Endianess and Alignment • Big Endian: address of most significant byte = word address (xx00 = Big End of word) • IBM 360/370, Motorola 68k, MIPS, Sparc, HP PA • Little Endian: address of least significant byte = word address(xx00 = Little End of word) • Intel 80x86, DEC Vax, DEC Alpha (Windows NT) little endian byte 0 3 2 1 0 lsb msb 0 1 2 3 0 1 2 3 Aligned big endian byte 0 Alignment: require that objects fall on address that is multiple of their size. Not Aligned

  46. Addressing Modes Meaning Addressing mode Example R4 ¬ R4+R3 Register Add R4,R3 R4 ¬ R4+3 Immediate Add R4,#3 R4 ¬ R4+Mem[100+R1] Displacement Add R4,100(R1) R4 ¬ R4+Mem[R1] Register indirect Add R4,(R1) R3 ¬ R3+Mem[R1+R2] Indexed / Base Add R3,(R1+R2) R1 ¬ R1+Mem[1001] Direct or absolute Add R1,(1001) R1 ¬ R1+Mem[Mem[R3]] Memory indirect Add R1,@(R3) R1 ¬ R1+Mem[R2]; R2 ¬ R2+d Post-increment Add R1,(R2)+ R2 ¬ R2–d; R1 ¬ R1+Mem[R2] Pre-decrement Add R1,–(R2) R1 ¬ R1+Mem[100+R2+R3*d] Scaled Add R1,100(R2)[R3] Why Post-increment/Pre-decrement? Scaled?

  47. register register MIPS Addressing Modes/Instruction Formats • All instructions 32 bits wide Register (direct) op rs rt rd Immediate immed op rs rt Base+index immed op rs rt Memory + PC-relative immed op rs rt Memory + PC

  48. Calls: Why Are Stacks So Great? Stacking of Subroutine Calls & Returns and Environments: A A: CALL B CALL C C: RET RET B: A B A B C A B A Some machines provide a memory stack as part of the architecture (e.g., VAX) Sometimes stacks are implemented via software convention (e.g., MIPS)

  49. Memory Stacks Useful for stacked environments/subroutine call & return even if operand stack not part of architecture Stacks that Grow Up vs. Stacks that Grow Down: 0 Little inf. Big Next Empty? Memory Addresses grows up grows down c b Last Full? a SP inf. Big 0 Little How is empty stack represented? Little  Big/Last Full POP: Read from Mem(SP) Decrement SP PUSH: Increment SP Write to Mem(SP) Little  Big/Next Empty POP: Decrement SP Read from Mem(SP) PUSH: Write to Mem(SP) Increment SP

  50. Call-Return Linkage: Stack Frames High Mem ARGS • Many variations on stacks possible (up/down, last pushed / next ) • Block structured languages contain link to lexically enclosing frame • Compilers normally keep scalar variables in registers, not memory! Reference args and local variables at fixed (positive) offset from FP Callee Save Registers (old FP, RA) Local Variables FP Grows and shrinks during expression evaluation SP Low Mem

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