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CPTR 220 Computer Organization Computer Architecture Assembly Programming 1 Chapter 3 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 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) 4 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 Why? Operands must be registers, only 32 registers provided – 32 bits per register. 32bits = Word – Names • $s0,$s1,.. For variables • $t0,$t1, .. For temporary variables • Design Principle: smaller is faster. Why? 5 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 Control Input Memory Datapath Processor Output I/O 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. 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 7 Memory Organization • • 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 8 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! 9 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 10 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 – See Figure 3.6 Page 121 for mapping of registers • Instruction Format: 000000 10001 op • rs 10010 rt 01000 rd 00000 100000 shamt funct Can you guess what the field names stand for? – Page 118 have the names for the field names 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) 35 op • • 18 8 rs rt 32 16 bit number Where's the compromise? See example on page 119 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 – Execute: Bits in the register "control" the subsequent actions – Fetch the “next” instruction and continue 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: .... 14 Control • MIPS unconditional branch instructions: j label • Example: if (i!=j) h=i+j; else h=i-j; beq $s4, $s5, Lab1 add $s3, $s4, $s5 j Lab2 Lab1: sub $s3, $s4, $s5 Lab2: ... 15 Control while (save[i] == k) i = i + j; Assume i = $s3; j = $s4; and k = $s5; save = $s6 Loop: add $t1,$s3,$s3 # $t1 = 2 * i add $t1,$t1,$t1 # $t1 = 4 * i add $t1, $t1 $s6 # $t1 = Addr(save[i]) lw $t0,0($t1) # $t0 = save[i] bne $t0,$s5, Exit # go to Exit if save[I] != k add $s3,$s3,$s4 #i=i+j j Loop Exit: …. 16 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 17 Control Flow • • We have: beq, bne, what about Branch-if-less-than? New instruction: if $s1 < $s2 then $t0 = 1 slt $t0, $s1, $s2 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 Register $Zero always contains a 0 MIPS does not include branch on less than because it is too complicated; either it would stretch the clock cycle time or this instruction would take extra clock cycles per instruction. • • • 18 2 Control Flow blt $s0,$s1,Label can be implemented as follows slt $t0,$s0,$s1 bne $t0,$zero, Label bgt $s0, $s1, Label can be implemented as follows slt $t0,$s1,$s0 bne $t0,$zero, Label 19 Control Flow • • • Another instruction – jr $s0 - would jump to the address stored in register $s0 Let us see the example on p129 Page 131 have all the instructions that we have learned so far. 20 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 21 Procedures • In order to implement procedures you must use the following instructions: – Jump-and-link to call a procedure • jal ProcedureName (or address) • jal places the return address in $ra – jr $ra to return from the procedure – You may need to save the contents of registers into memory by using a stack. • Use $sp as the stack pointer • Do not save $t# but only $s# 22 Procedure Example int leaf_example (int g, int h, int i, int j) { int f; f = (g + h ) – ( i + j); return f; } Leaf_example: subi $sp,$sp,4 sw $s0, 0($sp) add $t0,$a0,$a1 add $ti,$a2,$a3 sub $s0,$t0,$t1 add $v0, $s0, $zero lw $s0, 0($sp) addi $sp, $sp, 4 jr $ra 23 Nested Procedures • • • • • • • • • • • int fact (int n) { if (n < 1) return (1); else return (n * fact(n-1)); fact: subi $sp,$sp,8 #save $a0 = n and $ra sw $ra, 4($sp) sw $a0, 0($sp) slti $t0,$a0,1 beq $t0,$zero,L1 # if n >=1 go to L1 addi $v0,$zero,1 add $sp,$sp,8 # pop 2 items from the stack jr $ra 24 Factorial Procedure Cont. • • • • • • • • L1: subi $a0,$a0,1 # $a0 is n-1 jal fact # recursive call lw $a0, 0($sp) # restore old value of n and lw $ra, 4($sp) # return address addi $sp,$sp,8 mult $v0, $a0,$v0 jr $ra See page 141 for summary of all the instructions so far 25 Working with Strings • • • • • • • • • • • • • • • void strcpy (car x [ ], char y [ ]) { int i = 0; while ((x[i] = y [i]) != 0 ) i++; } strcpy: subi $sp,$sp,4 sw $s0,0($sp) add $s0,$zero,$zero L1: add $t1,$a1,$s0 lb $t2, 0($t1) add $t3,$a0,$s0 sb $t2, 0($t3) addi $s0,$s0,1 bne $t2,$zero,L1 lw $s0, 0($sp) addi $sp,sp,4 jr $ra # save $s0 # address of y[i] # load byte # address of x[i] 26 Constants • • Small constants are used quite frequently (50% of operands) e.g., A = A + 5; B = B + 1; C = C - 18; Solutions? – put 'typical constants' in memory and load them. Create hard-wired registers (like $zero) for constants like one. – Put the constants in the instructions • MIPS Instructions: addi $29, $29, 4 slti $8, $18, 10 andi $29, $29, 6 ori $29, $29, 4 • How do we make this work? – Use instruction format op(6),rs(5),rd(5),immediate(16) 27 3 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 28 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 29 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 30 Addresses in Branches and Jumps • • • Instructions: bne $t4,$t5,Label $t5 beq $t4,$t5,Label j Label Next instruction is at Label if $t4 != Next instruction is at Label if $t4 = $t5 Next instruction is at Label Formats: I op J op rs rt 16 bit address 26 bit address Addresses are not 32 bits 31 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) Since MIPS instructions are 4 bytes long, all PC-relative addressing refer to the number of words to the next instruction instead of the number of bytes. How many memory locations can you jump ? 32 Addressing • j L1 – Allows to jump to an address that is contained in a 26-bit field. – Because the number represent words so they could be stored in 28 bits – Use the upper 4 bits of PC to complete the 32 bits needed for addressing 33 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 34 Addressing Mode Summary • • • • • Register addressing – add $s1,$s2,$s3 Base or displacement addressing – lw $s0 8($t0) Immediate addressing – addi $s0, $s1, 5 PC-relative addressing – bne $s1, $s2,Exit Pseudodirect addressing – j L1 35 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 36 Decoding Machine Language • • See figure 3.18 See Example on page 154 37 Translation Steps • • • • • C Program -> Compiler -> Assembly Language Program -> Assembler -> Object Code -> Linker -> Executable - >Loader -> Execution 38 Procedure Swap 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 39 Procedure Sort sort (int v[], int n) { int i,j; for ( i = 0; i < n; i ++) for (j = i – 1; j > = 0 && v[j] > v [ j + 1]; j--) swap(v,j); } See MIPS assembly version of this sort on page 170 40 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 • Sometimes referred to as “RISC vs. CISC” – virtually all new instruction sets since 1982 have been RISC – VAX: minimize code size, make assembly language easy instructions from 1 to 54 bytes long! • We’ll look at PowerPC and 80x86 41 PowerPC • Indexed addressing – example: lw $t1,$a0+$s3 #$t1=Memory[$a0+$s3] – What do we have to do in MIPS? • • Update addressing – update a register as part of load (for marching through arrays) – example: lwu $t0,4($s3) #$t0=Memory[$s3+4];$s3=$s3+4 – What do we have to do in MIPS? Others: – load multiple/store multiple – a special counter register “bc Loop” decrement counter, if not 0 goto loop 42 80x86 • • • • • • 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: MMX is added “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” 43 A dominant architecture: 80x86 • • • See your textbook for a more detailed description 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” 44 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! 45