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chapter 2

Chapter 2. 2.1 IntroductionLanguage of the MachineWe'll be working with the MIPS instruction set architecturesimilar to other architectures developed since the 1980'sAlmost 100 million MIPS processors manufactured in 2002used by NEC, Nintendo, Cisco, Silicon Graphics, Sony,

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chapter 2

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    1. Chapter 2

    3. 2.2 Operations of the Computer Hardware All instructions have 3 operands Operand order is fixed (destination first) Example:

    4. MIPS arithmetic Design Principle: simplicity favors regularity. Of course this complicates some things...

    5. Example: C code: f = (g + h) (i + j); MIPS code: add t0, g, h add t1, i, j sub f, t0, t1

    6. 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

    7. 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

    8. 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.

    9. 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

    10. 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.

    11. Instructions Load and store instructions Example:

    12. Constant or Immediate Operands Constant in memory: lw $t0, AddrConstant4($s1) #$t0 = constant 4 add $s3, $s3, $t0 #$s3 = $s3 + $t0 Constant operand (add immediate): addi $s3, $s3, 4 #$s3 = $s3 + 4

    13. So far weve learned: MIPS loading words but addressing bytes arithmetic on registers only Instruction Meaning

    14. 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): 2.4 Representing Instructions in the Computer Board work: Binary NumbersBoard work: Binary Numbers

    15. 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) R-format and I-format

    16. 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

    17. So far weve learned:

    18. Instructions are bits Programs are stored in memory to be read or written just like data Stored Program Concept

    19. 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

    20. Logical opeartions: and or nor $t1: 0000 0000 0000 0000 0011 1100 0000 0000 $t2: 0000 0000 0000 0000 0000 1101 0000 0000 $t1 & $t2 0000 0000 0000 0000 0000 1100 0000 0000 $t1 | $t2 0000 0000 0000 0000 0011 1101 0000 0000 $t1: 0000 0000 0000 0000 0011 1100 0000 0000 $t3: 0000 0000 0000 0000 0000 0000 0000 0000 ~($t1 | $t3) 1111 1111 1111 1111 1100 0011 1111 1111 and $t0, $t1, $t2 # $t0 = $t1 & $t2 or $t0, $t1, $t2 # $t0 = $t1 | $t2 nor $t0, $t1, $t3 # $t0 = ~($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.

    21. So far weve learned:

    22. Decision making instructions alter the control flow, i.e., change the "next" instruction to be executed MIPS conditional branch instructions: 2.6 Instruction for Making Decisions

    23. MIPS unconditional branch instructions: j label Example: f =s0 g =s1 h =s2 i =s3 and j =s4 If-else

    24. 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 $t1, $s3, 2 add $t1, $t1, $s6 lw $$t0, 0($t1) bne $t0, $$5, Exit add $s3, $s3, 1 j loop Exit:

    25. So far: Instruction Meaning

    26. Continue set on less than instruction: slt $t0, $s3, $s4 slti $t0, $s2, 10 Means: if($s3<$s4) $t0=1; else $t0=0;

    27. 2.7 Supporting Procedures in Computer Hardware $a0-$a3: four arguments $v0-Sv1: two value registers in which to return values $ra: one return address register jal: jump and save return address in $ra. jr $ra: jump to the address stored in register $ra. Using More Registers Spill registers to memory Stack $sp

    28. Example: Leaf procedure int leaf_example(int g, int h, int i, int j) { int f; f=(g+h)-(i+j); return f; }

    29. 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

    30. 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

    31. Allocating Space for New Data on the Heap Need space for static variables and dynamic data structures (heap).

    32. Policy of Use Conventions

    33. 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

    34. 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 $s0, 0($sp) add $s0, $zero, $zero L1: add $t1, $s0, $a1 lb $t2, 0($t1) add $t3, $s0, $a0 sb $t2, 0($t3) beq $t2, $zero, L2 addi $s0, $s0, 1 j L1 L2: lw $s0, 0($sp) addi $sp, $sp, 4 jr $ra

    35. 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:

    36. 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

    37. Addressing in Branches and Jumps jump addressing j 10000 #go to location 10000

    38. Examples: Showing Branch Offset in Machine Language loop: sll $t1, $s3, 2 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit addi $s3, $s3, 1 j Loop Exit:

    39. 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: ...

    40. MIPS Addressing Modes Summary Register addressing Base or displacement addressing Immediate addressing PC-relative addressing Pseudodirect addressing: 26 bits concatenated with the upper bits of the PC.

    41. 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

    42. 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: Constants

    43. We'd like to be able to load a 32 bit constant into a register Must use two instructions, new "load upper immediate" instruction How about larger constants?

    44. 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 Assembly Language vs. Machine Language

    45. 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 Well talk about compiler optimizations when we hit chapter 4. Other Issues

    46. simple instructions all 32 bits wide very structured, no unnecessary baggage only three instruction formats Overview of MIPS

    47. 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 in Branches and Jumps

    48. 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 Formats: Addresses in Branches

    49. To summarize:

    51. 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 Alternative Architectures

    52. 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

    53. 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

    54. IA-32 Registers and Data Addressing Registers in the 32-bit subset that originated with 80386

    55. IA-32 Register Restrictions Registers are not general purpose note the restrictions below

    56. 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

    57. IA-32 instruction Formats Typical formats: (notice the different lengths)

    58. 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! Summary

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