Lecture 4 - Part 1

© Mahlke & Narayanasamy, 2008 The material in this presentation cannot be copied in any form without our written permission Prof. Scott Mahlke & Prof. Satish Narayanasamy EECS 370 ? Introduction to Computer Organization ? Winter 2008 EECS Department University of Michigan in Ann Arbor, USA 4. Instruction Set Architecture ? from C to assembly ? Part 1 EECS 370: Introduction to Computer Organization 2/31 The University of Michigan © 2008 From Last Time - MIPS Memory Instructions ? Load instructions ? lb \\ load byte signed (load 8 bits, sign extend) ? lbu \\ load byte unsigned (load 8 bits, zero extend) ? lh \\ load halfword (load 16 bits, sign extend) ? lhu \\ load halfword unsigned (load 16 bits, zero extend) ? lw \\ load word (load 32 bits, no extension) ? Store instructions (No sign/zero extension for stores) ? sb $3, 1000($4) \\ store 8 LSBs of $3 to M[$4+1000] ? sh $3, 1000($4) \\ store 16 LSBs of $3 to M[$4+1000] ? sw $3, 1000($4) \\ store all 32 bits of $3 to M[$4+1000] EECS 370: Introduction to Computer Organization 3/31 The University of Michigan © 2008 Big Endian vs Little Endian ? Endianness: ordering of bytes within a word ? Little - increasing numeric significance with increasing memory addresses ? Big ? The opposite, most significant byte first ? MIPS is Big endian EECS 370: Introduction to Computer Organization 4/31 The University of Michigan © 2008 From Last Time - Class Problem 3 What is the final state of memory once you execute the following instruction sequence? lhu $3, 100($0) lb $4, 102($0) sw $3, 100($0) sh $4, 102($0) 0xFF 0xFF 0x77 0xFF 100 101 102 103 $3 $4 register file Memory (each location is 1 byte) EECS 370: Introduction to Computer Organization 5/31 The University of Michigan © 2008 Instruction Set Architecture (ISA) Design Lectures ? Lecture 2: Storage types and addressing modes ? Lecture 3: LC-2K and MIPS architecture ? Lecture 4 (2 parts): Converting C to assembly: ? Data structures ? Branch instructions ? Calling functions / passing arguments ? Caller/callee save registers ? Lecture 5: Translation software; libraries, VMs EECS 370: Introduction to Computer Organization 6/31 The University of Michigan © 2008 Converting C to Assembly ? Memory layout ?memory addresses ? Branches ? Procedure calls ? Expression trees ? Register allocation EECS 370: Introduction to Computer Organization 7/31 The University of Michigan © 2008 Converting C to assembly ? example 1 Write MIPS Assembly Code for the following C Expression: C: a = b + names[ i ]; Assume that a is in $1, b is in $2, i is in $3, and the array names starts at address 1000 and holds 32-bit integers. lw $4, 1000 ($5) // load names[i] add $1, $2, $4 // calculate b + names[i] multi $5, $3, 4 // calculate array offset D A T A L A Y O U T EECS 370: Introduction to Computer Organization 8/31 The University of Michigan © 2008 Converting C to assembly ? example 2 Write MIPS Assembly Code for the following C Expression: class { int a; char b, c; } y; y.a = y.b + y.c; Assume that a pointer to y is in $1. lb $3, 5 ($1) // load y.c add $4, $2, $3 // calculate y.b+y.c lb $2, 4 ($1) // load y.b sw $4, 0 ($1) // store y.a D A T A L A Y O U T EECS 370: Introduction to Computer Organization 9/31 The University of Michigan © 2008 Calculating Load/Store Addresses for Variables short a[100]; char b; int c; double d; short e; struct { char f; int g[1]; char h; } i; Problem: Assume data memory starts at address 100, calculate the total amount of memory needed. a = 2 bytes * 100 = 200 b = 1 byte c = 4 bytes d = 8 bytes e = 2 bytes i = 1 + 4 + 1 = 6 bytes total = 221, right or wrong? D A T A L A Y O U T EECS 370: Introduction to Computer Organization 10/31 The University of Michigan © 2008 Memory layout of variables ? Cannot arbitrarily pack variables into memory ?Need to worry about alignment ? ?Golden? rule ? Address of a variable is aligned based on the size of the variable ? char is byte aligned (any addr is fine) ? short is half-word aligned (LSB of addr must be 0) ? int is word aligned (2 LSBs of addr must be 0) D A T A L A Y O U T EECS 370: Introduction to Computer Organization 11/31 The University of Michigan © 2008 Structure alignment ? Each field is laid out in the order it is declared using the Golden Rule for aligning ? Identify largest field ? Starting address of overall struct is aligned based on the largest field ? Size of overall struct is a multiple of the largest field ? Reason for this is so can have an array of structs D A T A L A Y O U T EECS 370: Introduction to Computer Organization 12/31 The University of Michigan © 2008 Structure Example struct { char w; int x[3] char y; short z; } The largest field is int (4 bytes), hence: struct size is multiple of 4 struct?s starting addr is word aligned Assume struct starts at location 1000, char w ?1000 x[0] ?1004-1007, x[1] ?1008 ? 1011, x[2] ?1012 ? 1015 char y ?1016 short z ?1018 ? 1019 Total size = 20 bytes! D A T A L A Y O U T EECS 370: Introduction to Computer Organization 13/31 The University of Michigan © 2008 Earlier Example ? 2 nd Try short a[100]; ?200 bytes ?100-299 char b; ? 1 byte ?300-300 int c; ? 4 bytes ? 304-307 double d; ?8 bytes ? 312-319 short e; ?2 bytes ?320-321 struct { ?largest field is 4 bytes ?start at 324 char f; ?1 byte ?324-324 int g[1]; ?4 bytes ?328 - 331 char h; ? 1 byte ?332-332 } i; ?struct size is 12 bytes, spanning 324 ? 335 236 bytes total!! Assume data memory starts at address 100 D A T A L A Y O U T EECS 370: Introduction to Computer Organization 14/31 The University of Michigan © 2008 Class Problem 1 ? How much memory is required for the following data, assuming that the data starts at address 200? int a; struct {double b, char c, int d} e; char *f; short g[20]; D A T A L A Y O U T EECS 370: Introduction to Computer Organization 15/31 The University of Michigan © 2008 MIPS Sequencing Instructions ? Sequencing instructions change the flow of instructions that are executed. ? This is achieved by modifying the program counter. ? Conditional branches ? If (condition_test) goto target_address - condition_test compares two operands (registers) - target_address is a 16-bit displacement on current PC+4 ? beq $x, $y, val - if ($x == $y) then PC = PC + 4 + val else PC = PC + 4 B R A N C H EECS 370: Introduction to Computer Organization 16/31 The University of Michigan © 2008 Setting the Displacement Field ? Target_address is a 16-bit displacement on current PC+4 ? Target = PC + 4 + displacement - beq $4, $0, 8 // branch 3 instr ahead if $4 = = 0 - beq $4, $0, -8 // branch 1 instruction back if $4 == 0 - beq $4, $0, -4 // Infinite loop if $4 == 0 add sub mult beq lw sw sll Example code sequence B R A N C H EECS 370: Introduction to Computer Organization 17/31 The University of Michigan © 2008 Other branching instructions ? bne $2, $3, offset // branch to offset+PC+4 if $2 ? $3 ? blt $2, $3, offset // branch to offset+PC+4 if $2 < $3 ? bge $2, $3, offset // pseudo-instruction $2 ? $3 ? j target // jump to target (pseudo-direct addressing mode) // uses a 26-bit target address concatenated to top 6 // bits of PC ? jr $3 // jump to address in $3 -- when is this useful? ? jal target // put PC+4 into register $31 and jump to target address B R A N C H EECS 370: Introduction to Computer Organization 18/31 The University of Michigan © 2008 Branch - example Convert the following C code into MIPS assembly (assume x is in $1, y in $2): int x, y; if (x == y) x++; else y++; ? bne $1, $2, 8 addi $1, $1, 1 beq $0, $0, 4 addi $2, $2, 1 Without Labels bne $1, $2, L1 addi $1, $1, 1 beq $0, $0, L2 L1: addi $2, $2, 1 L2: ? Using Labels Assemblers must deal with labels and assign displacements (L1) (L2) B R A N C H EECS 370: Introduction to Computer Organization 19/31 The University of Michigan © 2008 Loop - example // assume all variables are integers // i is in $1, start of a is 500, sum is in $2 for (i=0 ; i < 100 ; i++) { if (a[i] > 0) { sum += a[i]; } } add $1, $0, $0 addi $4, $0, 400 Loop1: bge $1, $4, endLoop lw $5, 500($1) ble $5, $0, endIf add $2, $2, $5 endIf: addi $1, $1, 4 j Loop1 endLoop: B R A N C H EECS 370: Introduction to Computer Organization 20/31 The University of Michigan © 2008 Converting function calls to assembly code C: printf(?hello world\n?); ? Need to pass parameters to the called function (printf) ? Need to save return address of caller ? Need to save register values ? Need to jump to printf ? Need to get return value (if used) ? Restore register values Execute instructions for printf() Jump to to return address F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 21/31 The University of Michigan © 2008 Task 1: Passing parameters ? Where should you put all of the parameters? ? Registers? - Fast access but few in number and wrong size for some objects ? Memory? - Good general solution but where? ? MIPS answer: ? Registers and memory. - Put the first few parameters in registers (if they fit) ($4 - $7) - Put the rest in memory on the call stack ? Example: add $4, $0, 1000 // put address of char array ?hello world? in $4 F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 22/31 The University of Michigan © 2008 Call stack ? MIPS conventions (and most other processors) allocate a region of memory for the call stack ? This memory is used to manage all the storage requirements to simulate function call semantics - Parameters (that were not passed through registers) - Local variables - Temporary storage (when you run out of registers and need somewhere to save a value) - Return address - Etc. ? Sections of memory on the call stack [a.k.a. stack frames] are allocated when you make a function call, and de-allocated when you return from a function. F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 23/31 The University of Michigan © 2008 MIPS Memory Map Text (instructions) Static data stack heap 7fff ffff 1000 0000 1000 8000 0040 0000 Stack: starts at 0x7fff ffff and grows down to lower addresses. Bottom of the stack resides in the $sp register Heap: starts at 0x1000 8000 and grows up to higher addresses. Allocation done explicitly with malloc(). Deallocation with free(). Runtime error if no free mem before running into $sp address. Static: starts at 0x1000 0000. Holds all global variables and those locals explicitly declared as ?static?. Text: starts at 0x0040 0000. Holds all instructions in the program (except for Dynamically linked library routines DLLs) frames F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 24/31 The University of Michigan © 2008 The MIPS Stack Frame Return Address Incoming parameters Spilled Registers Local Vars Outgoing parameters $SP $FP static text stack heap Frame $SP $FP F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 25/31 The University of Michigan © 2008 Allocating space to local variables ? Local variables (by default) are created when you enter a function, and disappear when you leave ? Technical terminology: local variables are placed in the automatic storage class (as opposed to the static storage class used for globals). ? Automatics are allocated on the call stack ? How? by incrementing (or decrementing) the pointer to the top of the call stack ? sub $sp, $sp, 12 // allocate space for 3 integer locals add $sp, $sp, 12 // de-allocate space for locals F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 26/31 The University of Michigan © 2008 The stack grows as functions are called void foo() { int x, y[2]; bar(x); } void bar(int x) { int a[3]; printf(); } foo?s stack frame inside foo foo calls bar foo?s stack frame bar?s stack frame bar calls printf foo?s stack frame bar?s stack frame printf?s stack frame F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 27/31 The University of Michigan © 2008 The stack shrinks as functions return void foo() { int x, y[2]; bar(x); } void bar(int x) { int a[3]; printf(); } foo?s stack frame bar returns printf returns foo?s stack frame bar?s stack frame F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 28/31 The University of Michigan © 2008 Stack frame contents x return addr to main y[0] spilled regs in foo y[1] inside foo ? foo?s stack frame void foo() { int x, y[2]; bar(x); } void bar(int x) { int a[3]; printf(); } F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 29/31 The University of Michigan © 2008 Stack frame contents (2) x return addr to foo a[0] a[1] a[2] spilled regs in bar foo?s f rame bar?s frame void foo() { int x, y[2]; bar(x); } void bar(int x) { int a[3]; printf(); } x return addr to main y[0] spilled regs in foo y[1] foo calls bar F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 30/31 The University of Michigan © 2008 Stack frame contents (3) printf local vars return addr to bar printf?s frame void foo() { int x, y[2]; bar(x); } void bar(int x) { int a[3]; printf(); } x return addr to foo a[0] a[1] a[2] spilled regs in bar foo?s f rame bar?s frame x return addr to main y[0] spilled regs in foo y[1] bar calls printf F U N C T I O N C A L L S EECS 370: Introduction to Computer Organization 31/31 The University of Michigan © 2008 Assigning variables to memory spaces int w; void foo(int x) { static int y[4]; char *p; p = malloc(10); ? printf(?%s\n?, p); } static text stack heap w goes in static, as it?s a global x goes on the stack, as it?s a parameter y goes in static, 1 copy of this!! p goes on the stack allocate 10 bytes on heap, ptr set to the address string goes in static, pointer to string on stack, p goes on stack Gary Tyson 370 lecture 4 MIPS ISA concl, vars in C, function calls (begin)