Lecture 8 Sept 23 Completion of Ch 2 translating procedure into MIPS role of compilers, assemblers, linking, loading etc. pitfalls, conclusions Chapter.

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Presentation transcript:

Lecture 8 Sept 23 Completion of Ch 2 translating procedure into MIPS role of compilers, assemblers, linking, loading etc. pitfalls, conclusions Chapter 3 ?

Procedure Calling Steps required 1.Place parameters (arguments) in registers 2.Transfer control to procedure 3.Acquire storage for procedure 4.Perform procedure’s operations 5.Place result in register for caller 6.Return to place of call §2.8 Supporting Procedures in Computer Hardware

Register Usage $a0 – $a3: arguments (reg’s 4 – 7) $v0, $v1: result values (reg’s 2 and 3) $t0 – $t9: temporaries Can be overwritten by callee $s0 – $s7: saved Must be saved/restored by callee $gp: global pointer for static data (reg 28) $sp: stack pointer (reg 29) $fp: frame pointer (reg 30) $ra: return address (reg 31)

Procedure Call Instructions Procedure call: jump and link jal ProcedureLabel Address of following instruction put in $ra Jumps to target address Procedure return: jump register jr $ra Copies $ra to program counter Can also be used for computed jumps e.g., for case/switch statements

Leaf Procedure Example c code: int leaf_example (int g, h, i, j) { int f; f = (g + h) - (i + j); return f; } Arguments g, …, j in $a0, …, $a3 f in $s0 (hence, need to save $s0 on stack) Result in $v0

Leaf Procedure Example MIPS code: leaf_example: addi $sp, $sp, -4 sw $s0, 0($sp) add $t0, $a0, $a1 add $t1, $a2, $a3 sub $s0, $t0, $t1 add $v0, $s0, $zero lw $s0, 0($sp) addi $sp, $sp, 4 jr $ra Save $s0 on stack Procedure body Restore $s0 Result Return

Non-Leaf Procedures Procedures that call other procedures are known as non-leaf procedures. For nested call, caller needs to save on the stack: Its return address Any arguments and temporaries needed after the call Restore from the stack after the call

Non-Leaf Procedure Example C code: int fact (int n) { if (n < 1) return 1; else return n * fact(n-1); } Argument n in $a0 Result in $v0

Non-Leaf Procedure Example MIPS code: fact: addi $sp, $sp, -8 # adjust stack for 2 items sw $ra, 4($sp) # save return address sw $a0, 0($sp) # save argument slti $t0, $a0, 1 # test for n < 1 beq $t0, $zero, L1 addi $v0, $zero, 1 # if so, result is 1 addi $sp, $sp, 8 # pop 2 items from stack jr $ra # and return L1: addi $a0, $a0, -1 # else decrement n jal fact # recursive call lw $a0, 0($sp) # restore original n lw $ra, 4($sp) # and return address addi $sp, $sp, 8 # pop 2 items from stack mul $v0, $a0, $v0 # multiply to get result jr $ra # and return

As we saw in the previous slide, the program correctly computes 12! As 12! = But when tried for n = 13, the output is: This is clearly wrong! What happened? How to fix the program?

Memory Layout Text: program code Static data: global variables e.g., static variables in C, constant arrays and strings $gp initialized to address allowing ±offsets into this segment Dynamic data: heap E.g., malloc in C, new in Java Stack: automatic storage

Byte/Halfword Operations Could use bitwise operations MIPS byte/halfword load/store String processing is a common case lb rt, offset(rs) lh rt, offset(rs) Sign extend to 32 bits in rt lbu rt, offset(rs) lhu rt, offset(rs) Zero extend to 32 bits in rt sb rt, offset(rs) sh rt, offset(rs) Store just rightmost byte/halfword

Target Addressing Example Loop code from earlier example Assume Loop at location Loop: sll $t1, $s3, add $t1, $t1, $s lw $t0, 0($t1) bne $t0, $s5, Exit addi $s3, $s3, j Loop Exit: … 80024

Addressing Mode Summary

Translation and Startup Many compilers produce object modules directly Static linking §2.12 Translating and Starting a Program

Assembler Pseudoinstructions Most assembler instructions represent machine instructions one-to-one Pseudoinstructions: not supported in instruction set, but assembler translates into equivalent move $t0, $t1 → add $t0, $zero, $t1 blt $t0, $t1, L → slt $at, $t0, $t1 bne $at, $zero, L $at (register 1): assembler temporary

Assembler Pseudoinstructions Another example: la $r0, label Gets translated into MIPS as follows: Suppose the address of label is ABCD (in hex) lui $r0, 0xAB addi $r0, 0xCD

Producing an Object Module Assembler (or compiler) translates program into machine instructions Provides information for building a complete program from the pieces Header: described contents of object module Text segment: translated instructions Static data segment: data allocated for the life of the program Relocation info: for contents that depend on absolute location of loaded program Symbol table: global definitions and external refs Debug info: for associating with source code

Linking Object Modules Produces an executable image 1.Merges segments 2.Resolve labels (determine their addresses) 3.Patch location-dependent and external refs

Loading a Program Load from image file on disk into memory 1.Read header to determine segment sizes 2.Create virtual address space 3.Copy text and initialized data into memory Or set page table entries so they can be faulted in 4.Set up arguments on stack 5.Initialize registers (including $sp, $fp, $gp) 6.Jump to startup routine Copies arguments to $a0, … and calls main When main returns, do exit syscall

Dynamic Linking Only link/load library procedure when it is called Requires procedure code to be relocatable Avoids image bloat caused by static linking of all (transitively) referenced libraries Automatically picks up new library versions

ARM & MIPS Similarities ARM: the most popular embedded core Similar basic set of instructions to MIPS §2.16 Real Stuff: ARM Instructions ARMMIPS Date announced1985 Instruction size32 bits Address space32-bit flat Data alignmentAligned Data addressing modes93 Registers15 × 32-bit31 × 32-bit Input/outputMemory mapped

Compare and Branch in ARM Uses condition codes for result of an arithmetic/logical instruction Negative, zero, carry, overflow Compare instructions to set condition codes without keeping the result Each instruction can be conditional Top 4 bits of instruction word: condition value Can avoid branches over single instructions

Instruction Encoding

Fallacies Powerful instruction  higher performance Fewer instructions required But complex instructions are hard to implement May slow down all instructions, including simple ones Compilers are good at making fast code from simple instructions Use assembly code for high performance But modern compilers are better at dealing with modern processors More lines of code  more errors and less productivity §2.18 Fallacies and Pitfalls

Concluding Remarks Design principles 1.Simplicity favors regularity 2.Smaller is faster 3.Make the common case fast 4.Good design demands good compromises Layers of software/hardware Compiler, assembler, hardware MIPS: typical of RISC ISAs §2.19 Concluding Remarks

Concluding Remarks Measure MIPS instruction executions in benchmark programs Consider making the common case fast Consider compromises Instruction classMIPS examplesSPEC2006 IntSPEC2006 FP Arithmetic add, sub, addi 16%48% Data transfer lw, sw, lb, lbu, lh, lhu, sb, lui 35%36% Logical and, or, nor, andi, ori, sll, srl 12%4% Cond. Branch beq, bne, slt, slti, sltiu 34%8% Jump j, jr, jal 2%0%