Istruzioni: linguaggio macchina. Università di Bergamo - corso di Calcolatori Elettronici

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1 Istruzioni: linguaggio macchina 1

2 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 2

3 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) 3

4 MIPS arithmetic Design Principle: simplicity favors regularity. Why? 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 Hp: A $s0, B $s1, C $s2, D $s3, E $s4, F $s5 Operands must be registers, only 32 registers provided Design Principle: smaller is faster. Why? 4

5 Registers vs. Memory Arithmetic instructions operands must be registers, only bit registers provided (word in MIPS architecture) Compiler associates variables with registers What about programs with lots of variables Control Input Memory Datapath Output Processor I/O 5

6 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 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 6

7 Memory Organization Bytes are nice, but most data items use larger "words" For MIPS, a word is 32 bits or 4 bytes bits of data 32 bits of data 32 bits of data 32 bits of data Registers hold 32 bits of data 2 32 bytes with byte addresses from 0 to words with byte addresses 0, 4, 8, Words are aligned i.e., what are the least 2 significant bits of a word address? 7

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

9 So far we ve learned MIPS loading words but addressing bytes arithmetic on registers only Instruction Meaning add $s1, $s2, $s3 $s1 = $s2 + $s3 sub $s1, $s2, $s3 $s1 = $s2 $s3 lw $s1, 100($s2) $s1 = Memory[$s2+100] sw $s1, 100($s2) Memory[$s2+100] = $s1 9

10 Machine Language 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: op rs rt rd shamt funct Can you guess what the field names stand for? 10

11 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) op rs rt 16 bit number Where's the compromise? 11

12 Stored Program Concept Instructions are bits Programs are stored in memory to be read or written just like data Processor Memory 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 12

13 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:... Hp: i $s0, j $s1, h $s3 13

14 Control if then else 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:... Hp: h $s3, i $s4, j $s5 Can you build a simple for loop? 14

15 Control Loops E importante disporre di un istruzione per iterare: Esempio: Loop:g=g+A[i]; Loop: add $t1, $s3, $s3 i=i+j; add $t1, $t1, $t1 if (i h) goto Loop add $t1, $t1, $s5 lw $t0, 0($t1) add $s1, $s1, $t0 add $s3, $s3, $s4 bne $s3, $s2, Loop BASIC BLOCK Hp: g $s1, h $s2, i $s3, j $s4, A $s5, 15

16 Control Loop C code: L1: g = g + A[i]; i = i + j; if (i!= h) goto L1; MIPS code: L1: add $t1, $s3, $s3 # $t1 = 2 * i add $t1, $t1, $t1 # $t1 = 4 * i add $t1, $t1, $s5 # $t1 = indirizzo di A[i] lw $t0, 0($t1) # $t0 = A[i] add $s1, $s1, $t0 # g = g + A[i] add $s3, $s3, $s4 # i = i + j bne $s3, $s2, L1 # vai a L1 se i h Attribuzione dei registri alle variabili: g = $s1, h = $s2, i = $s3, j = $s4, indirizzo di inizio del vettore A = $s5 16

17 Control While loop Più comodo disporre di un istruzione while che ciclare con goto: Esempio: while (save [i] ==k) Loop: add $t1, $s3, $s3 i=i+j; add $t1, $t1, $t1 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit add $s3, $s3, $s4 j Loop Exit:.. Hp: i $s3, j $s4, k $s5, save $s6 17

18 Ciclo While C code: while (save [i] == k) i = i + j; MIPS code: Loop: add $t1, $s3, $s3 add $t1, $t1, $t1 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit add $s3, $s3, $s4 j Loop Exit: # $t1 = 2 * i # $t1 = 4 * i # $t1 = indirizzo di save[i] # $t0 = save[i] # vai a Exit se save[i] k # i = i + j # vai a Loop Attribuzione dei registri alle variabili: i = $s3, j = $s4, k = $s5, indirizzo di inizio del vettore save = $s6 18

19 Control Flow We have: beq, bne, what about Branch-if-less-than? New instruction: slt $t0, $s1, $s2 if $s1 < $s2 then $t0 = 1 else $t0 = 0 Can use this instruction to build "blt $s1, $s2, Label" can now build general control structures Note that the assembler needs a register to do this, there are policy of use conventions for registers bne $t0, $zero, less slt+bne = branch on less (less complicated than a specific instruction) 19

20 Control Case/Switch Selezione tra diverse alternative a seconda del valore di una variabile: anziché n if then else, una tabella indicizzata (jump address table) contenente indirizzi che corrispondono a etichette nel codice jr salto incondizionato a un registro switch(k) slt $t3, $s5, $zero case 0 f=i+j;break; bne $t3, $zero, Exit case 1 f=g+h;break; slt $t3, $s5, $t2 case 2 f=g-h;break; beq $t3, $zero, Exit case 3 f=i-j;break; add $t1, $s5, $s5 add $t1, $t1, $t1 add $t1, $t1, $t4 lw $t0, 0($t1) jr $t0 L0: add $s0, $s3, $s4; j Exit L1: add $s0, $s1, $s2; j Exit L2: sub $s0, $s1, $s2; j Exit L3: sub $s0, $s3, $s4; j Exit Exit:.. Hp: f $s0, g $s1, h $s2, i $s3, j $s4, k $s5, 4 $t2, indirizzo tabella $t4 20

21 Case/Switch Switch (k) { slt $t3, $s5, $zero # k <0? case 0: f=i+j; break; bne $t3, $zero, Exit case 1: f=g+h; break; slt $t3, $s5, $t2 # k >3? case 2: f=g-h; break; beq $t3, $zero, Exit case 3: f=i-j; break; add $t1, $s5, $s5 } add $t1, $t1, $t1 # $t1=4*k add $t1, $t1, $t4 lw $t0, 0($t1) jr $t0 #vai a indir. letto L0: add $s0, $s3, $s4 #k=0, f=i+j j Exit L1: add $s0, $s1, $s2 #k=1, f=g+h j Exit L2: sub $s0, $s1, $s2 #k=2, f=g-h j Exit L3: sub $s0, $s3, $s4 #k=3, f=i-j Exit: f = $s0, g = $s1, h = $s2, i = $s3, j = $s4, k = $s5; $t2 = 4; $t4 =indirizzo tabella etichette 21

22 So far Instruction Meaning add $s1,$s2,$s3 $s1 = $s2 + $s3 sub $s1,$s2,$s3 $s1 = $s2 $s3 lw $s1,100($s2) $s1 = Memory[$s2+100] sw $s1,100($s2) Memory[$s2+100] = $s1 bne $s4,$s5,l Next instr. is at Label if $s4 $s5 beq $s4,$s5,l Next instr. is at Label if $s4 = $s5 j Label Next instr. is at Label Formats: R I J op rs rt rd shamt funct op rs rt 16 bit address op 26 bit address 22

23 Policy of Use Conventions Name Register number Usage $zero 0 the constant value 0 $v0-$v1 2-3 values for results and expression evaluation $a0-$a3 4-7 arguments $t0-$t temporaries $s0-$s saved $t8-$t more temporaries $gp 28 global pointer $sp 29 stack pointer $fp 30 frame pointer $ra 31 return address 23

24 Uso dello stack: procedure foglia $sp=registro stack pointer: punta all ultima locazione occupata dello stack N.B. indirizzi crescenti verso l alto: pushsp-4, pop sp+4 int proc(int g, int h, int i, int j) sub $sp, $sp, 12 { int f; sw $t1, 8($sp) f=(g+h)+(i+j); sw $t0, 4($sp) return(f); sw $s0, 0($sp) add $t0, $a0, $a1 } g, h, i, j $a0 $a3 add $t1, $a2, $a3 f $s0 add $s0, $t0, $t1 add $v0, $s0, $zero Per convenzione: lw $s0, 0($sp) $t0-$t9 temporanei da non salvare $s0-$s7 da conservare si potevano risparmiare 2 push/pop lw $t0, 4($sp) lw $t1, 8($sp) add $sp, $sp, 12 jr $ra 24

25 Uso dello stack int proc (int g, int h, int i, int j) { int f; f=(g+h)-(i+j); return f; proc: addi $sp, $sp, -12 # 3 push sw $t1, 8($sp) } sw $t0, 4($sp) g, h, i, j = $a0 $a3 sw $s0, 0($sp) f = $s0 add $t0, $a0, $a1 # calc. f add $t1, $a2, $a3 sub $s0, $t0, $t1 add $v0, $s0, $zero # $v0=f lw $s0, 0($sp) # 3 pop lw $t0, 4($sp) lw $t1, 8($sp) addi $sp, $sp, 12 jr $ra # ritorno 25

26 Uso dello stack: procedure annidate Pb: procedure che ne chiamano altreconflitto nel passaggio parametri? Sol.: uso dello stack: chiamante salva $a0-$a3 o $t0-$t9 usati dal chiamato; chiamato salva $ra e $s0-$s7 usati dal chiamante. int fact(int n) fact sub $sp, $sp, 8 { sw $ra, 4($sp) if n<1 return(1); sw $a0, 0($sp) else return(n fact(n-1)); slt $t0, $a0, 1 } beq $t0, $zero, L1 n $a0 add $v0, $zero, 1 add $sp, $sp, 8 jr $ra L1: sub $a0, $a0, 1 jal fact lw $a0, 0($sp) lw $ra, 4($sp) add $sp, $sp, 8 mult $v0, $a0, $v0 jr $ra 26

27 Procedure annidate int fatt (int n) fatt: addi $sp, $sp, -8 { sw $ra, 4($sp) if (n<1) return(1); sw $a0, 0($sp) else return(n*fatt(n-1)); slti $t0, $a0, 1 } beq $t0, $zero, L1 addi $v0, $zero, 1 addi $sp, $sp, 8 n = $a0 jr $ra L1: addi $a0, $a0, -1 jal fatt lw $a0, 0($sp) # ind. = L1+8 lw $ra, 4($sp) addi $sp, $sp, 8 mult $v0, $a0, $v0 jr $ra 27

28 Procedure annidate Gestione dello stack: Al 1 richiamo salva nello stack: 1) l indirizzo di ritorno che è nella zona del chiamante (nome attribuito JALM + 4); 2) il valore di $a0 = n. Al 2 richiamo salva nello stack: 1) l indirizzo della procedura fatt (indicato da L1+8); 2) il valore di $a0 = n-1. Al 3 richiamo salva nello stack L1+8 e $a0 = n Al n-mo richiamo salva nello stack L1+8 e $a0 = 0. 28

29 Procedure annidate Esempi di esecuzione al variare di n: n = 0 $ra = JALM+4 $a0 = n = 0 n = 1 $ra = JALM+4 $a0 = n = 1 $ra = L1+8 $a0 = n-1 = 0 1^ esecuzione 2^ esecuzione Alla 1^ iterazione: salta a L1; $a0 = 0; $ra=l1+8. Alla 2^ iterazione: non salta a L1; $v0=1 e ritorna a L1+8, dove $a0=1; $ra=jalm+4; $v0=$v0*1 e ritorna al main. 29

30 Procedure annidate Esempi di esecuzione al variare di n: n = 2 $ra = JALM+4 $a0 = n = 2 $ra = L1+8 $a0 = n-1 = 1 $ra = L1+8 $a0 = n-2 = 0 1^ esecuzione 2^ esecuzione 3^ esecuzione Alla 1^ iterazione: salta a L1; $a0 diventa 1; $ra=l1+8. Alla 2^ iterazione: salta a L1; $a0 diventa 0; $ra=l1+8. Alla 3^ iterazione: non salta a L1, quindi $v0=1 e torna a L1+8, $a0=1; $ra=l1+8; $v0=$v0*1; torna a L1+8, $a0=2, $ra=jalm+4, $v0=1*$a0=2 e torna al main program. 30

31 Procedure annidate fatt. Salva indirizzo di ritorno e valore a0 nello stack Richiamo fatt sì a0<1 no v0=1 dec a0 Preleva a0 Ultima iterazione Ritorno Ritorno all ultima chiamata effettuata (2 casi: n-1 volte si ritorna alla routine fatt. all indirizzo L1+8 e si preleva a0 dallo stack, solo l ultima si torna al main (JALM+4)) e si aggiorna SP Iter. intermedie v0=a0*v0 Fatto con il ritorno alla routine fatt che aveva chiamato 31

32 Fattoriale senza ricorsione n nello stack FATT=1 sì a0<2 no n dallo stack FATT=FATT*a0 a0=a0-1 ritorno 32

33 Fattoriale senza ricorsione n = $a0 fatt: addi $sp, $sp, -4 # agg. $sp per salvat. n sw $a0, 0($sp) # salvataggio n addi $v0, $zero, 1 # $v0 = fattoriale = 1 Ciclo: slti $t0, $a0, 2 # test per $a0 < 2 beq $t0, $zero, L1 # salta se $a0 >= 2 lw $a0, 0($sp) # ripristino n addi $sp, $sp, 4 # aggiornamento $sp jr $ra L1: mult $v0, $a0, $v0 # fatt = fatt * $a0 addi $a0, $a0, -1 # $a0 = $a0-1 j Ciclo 33

34 Uso stack per variabili locali alle procedure Esempio: variabili non memorizzabili in registri (array, strutture) procedure frame/activation record: segmento stack contenente registri salvati e variabili locali $fp= frame pointer (puntatore all inizio del frame, comodo perché è un base register stabile nel frame, mentre $sp può cambiare) $fp $a0-$a3 $fp $fp $sp $ra $sp $s0 $s7 Local var. $sp N.B. Se si passano più di 4 parametri, i primi in $a0-$a3, gli altri direttamente nello stack 34

35 Gestione caratteri void strcpy (char [x], char [y]) { int i; i = 0; while ((x[i] = y[i])!= \0 ) /* copia e test byte */ i = i + 1; } strcpy: addi $sp, $sp, -4 sw $s0, 0($sp) # salva $s0 nello stack add $s0, $zero, $zero # i = 0 L1: add $t1, $a1, $s0 # ind. y[i] in $t1 lb $t2, 0($t1) # $t2 = y[i] add $t3, $a0, $s0 # ind. x[i] in $t3 sb $t2, 0($t3) # x[i] = y[i] addi $s0, $s0, 1 # i = i + 1 bne $t2, $zero, L1 # se y[i] 0 vai a L1 lw $s0, 0($sp) # ripristina $s0 dallo stack addi $sp, $sp, 4 jr $ra # ritorno Indirizzo stringa x = $a0; ind. Y = $a1; i = $s0 35

36 Another example Can we figure out the code? swap(int v[], int k); { int temp; temp = v[k] v[k] = v[k+1]; v[k+1] = temp; } swap: add $t1, $a1, $a1 add $t1, $t1, $t1 add $t1, $a0, $t1 lw $t0, 0($t1) lw $t2, 4($t1) sw $t2, 0($t1) sw $t0, 4($t1) jr $ra 36

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. create hard-wired registers (like $zero) for constants like one. MIPS Instructions: addi $sp, $sp, 4 slti $t0, $s3, 10 andi $sp, $sp, 6 ori $sp, $sp, 4 How do we make this work? 37

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

39 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 39

40 Pseudo Istruzioni Versioni modificate delle istruzioni vere, trattate dall assemblatore Esempi: Pseudo istruzione: move $t0, $t1 # $t0 = $t1 Istruzione vera: add $t0, $zero, $t1 Altri esempi: Pseudo istruzione: blt $s1, $s2, Label Istruzioni vere: slt $at, $s1, $s2 bne $at, $zero, Label bgt, bge, ble; branch condizionati a locazioni distanti trasformati in un branch e una jump, li, etc. 40

41 Other Issues Things we are not going to cover 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 about later We've focused on architectural issues basics of MIPS assembly language and machine code we ll build a processor to execute these instructions. 41

42 Overview of MIPS simple instructions all 32 bits wide very structured, no unnecessary baggage only three instruction formats R I op rs rt rd shamt funct op rs rt 16 bit address J op 26 bit address rely on compiler to achieve performance what are the compiler's goals? help compiler where we can 42

43 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: I J op rs rt 16 bit address op 26 bit address Addresses are not 32 bits How do we handle this with load and store instructions? 43

44 Addresses in Branches Instructions: bne $t4,$t5,label beq $t4,$t5,label Formats: Next instruction is at Label if $t4 $t5 Next instruction is at Label if $t4=$t5 I 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 44

45 To summarize MIPS operands Name Example Comments $s0-$s7, $t0-$t9, $zero, Fast locations for data. In MIPS, data must be in registers to perform 32 registers $a0-$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], Accessed only by data transfer instructions. MIPS uses byte addresses, so 2 30 memory Memory[4],..., sequential words differ by 4. Memory holds data structures, such as arrays, words Memory[ ] and spilled registers, such as those saved on procedure calls. MIPS assembly language Category Instruction Example Meaning Comments add add $s1, $s2, $s3 $s1 = $s2 + $s3 Three operands; data in registers Arithmetic subtract sub $s1, $s2, $s3 $s1 = $s2 - $s3 Three operands; data in registers add immediate addi $s1, $s2, 100 $s1 = $s Used to add constants load word lw $s1, 100($s2) $s1 = Memory[$s ] Word from memory to register store word sw $s1, 100($s2) Memory[$s ] = $s1 Word from register to memory Data transfer load byte lb $s1, 100($s2) $s1 = Memory[$s ] Byte from memory to register store byte sb $s1, 100($s2) Memory[$s ] = $s1 Byte from register to memory load upper immediate lui $s1, 100 $s1 = 100 * 2 16 Loads constant in upper 16 bits branch on equal beq $s1, $s2, 25 if ($s1 == $s2) go to PC branch on not equal bne $s1, $s2, 25 if ($s1!= $s2) go to Conditional PC branch set on less than slt $s1, $s2, $s3 if ($s2 < $s3) $s1 = 1; else $s1 = 0 Equal test; PC-relative branch Not equal test; PC-relative Compare less than; for beq, bne set less than immediate slti $s1, $s2, 100 if ( $s2 < 100) $s1 = 1; else $s1 = 0 Compare less than constant jump j 2500 go to Jump to target address Uncondi- jump register jr $ra go to $ra For switch, procedure return tional jump jump and link jal 2500 $ra = PC + 4; go to For procedure call 45

46 Modi di indirizzamento Register addressing operando = registro Base addressing operando = cost($reg) Immediate addressing operando = costante PC relative addressing operando = PC+cost. Pseudodirect addressing operando = 4 bit PC + 26 istr. 46

47 1. Immediate addressing op rs rt Immediate 2. Register addressing op rs rt rd... funct Registers Register 3. Base addressing op rs rt Address Memory Register + Byte Halfword Word 4. PC-relative addressing op rs rt Address Memory PC + Word 5. Pseudodirect addressing op Address Memory PC Word 47

48 Vettori e puntatori Clear1(int array[], int size) { int i; for (i=0; i<size; i++) array[i]=0; } clear2(int *array, int size) { int *p; // *p= oggetto puntato da p for (p=&array[0]; p< &array[size]; p++) //&...= indirizzo... *p=0; } 48

49 Vettori e puntatori azz1 (int vett[], int dim) { int i; for (i=0; i<dim; i++) vett[i] = 0; } azz1: move $t0, $zero # i = 0 L1: add $t1, $t0, $t0 # 4 * i add $t1, $t1, $t1 add $t2, $a0, $t1 # $t2 = indirizzo di vett[i] sw $zero, 0($t2) # vett[i] = 0 addi $t0, $t0, 1 # i = i + 1 slt $t3, $t0, $a1 # i < dim? bne $t3, $zero, L1 # se i < dim vai a L1 jr $ra Indirizzo vett = $a0, dim = $a1, i = $t0 49

50 Vettori e puntatori azz2 (int *vett, int dim) { // *p è l oggetto puntato da p int *p; // &vett è l indirizzo di vett for (p=&vett[0]; p<&vett[dim]; p++) *p = 0; } azz2: move $t0, $a0 # p = indir vett[0] add $t1, $a1, $a1 # 4 * dim add $t1, $t1, $t1 add $t2, $a0, $t1 # $t2 = indir di vett[dim] L2: sw $zero, 0($t0) # mem puntata da p = 0 addi $t0, $t0, 4 # p = p + 4 slt $t3, $t0, $t2 # p < &vett[dim]? bne $t3, $zero, L2 # se è vero vai a L2 jr $ra Indirizzo vett = $a0, dim = $a1, p = $t0 50

51 Vettori e puntatori Clear1 Clear2 move $t0,$zero move $t0,$a0 loop1: add $t1,$t0,$t0 add $t1,$a1,$a1 add $t1,$t1,$t1 add $t1,$t1,$t1 add $t2,$a0,$t1 add $t2,$a0,$t1 sw $zero, 0($t2) loop2: sw $zero,0($t0) addi $t0,$t0,1 addi $t0,$t0,4 slt $t3, $t0, $a1 slt $t3,$t0,$t2 bne $t3, $zero, loop1 bne $t3,$zero,loop2 Hp. array $a0, size $a1, i $t0 array $a0, size $a1, p $t0 multiply and add all interno del loop solo incremento di p guadagno 7 su 4 (1.75) 51

52 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! 52