Computer Architecture Organization Instructions PCSpim Tutorial Engr Umbreen

Computer Architecture & Organization Instructions: PCSpim Tutorial Engr. Umbreen Sabir Computer Engineering Department, University of Engg. & Technology Taxila. 1 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Adventures in Assembly Land l l l 2 What is an Assembler ASM Directives ASM Syntax Intro to PCSPIM Simple examples 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

A Simple Programming Task l l 3 Add the numbers 0 to 4 … 10 = 0 + 1 + 2 + 3 + 4 In “C”: int i, sum; main() { sum = 0; for (i=0; i<5; i++) sum = sum + I; } Now let’s code it in ASSEMBLY 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

What IS an Assembler? l A program for writing programs l Assembler: 1. A Symbolic LANGUAGE for representing strings of bits. 2. A PROGRAM for translating Assembly Source to binary. 4 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Assembly Source Language l l l 5 l An Assembly SOURCE FILE contains, in symbolic text, values of successive bytes to be loaded into memory. . . e. g. . data 0 x 10000000 Specifies address where following values are loaded. byte 1, 2, 3, 4 #First four byte values. byte 5, 6, 7, 8 # Second four byte values. word 1, 2, 3, 4 # 4 WORD values (each is 4 bytes). asciiz "Comp 411“ # A string of 9 ASCII bytes. 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Assembly Source Language cont. l l l . align 2 # Align to next multiple of 4 (22). word 0 xfeedbeef # Hex encoded WORD Value Resulting memory dump: [0 x 10000000] 0 x 04030201 0 x 08070605 0 x 00000001 0 x 00000002 [0 x 10000010] 0 x 00000003 0 x 00000004 0 x 706 d 6 f 43 0 x 31313420 [0 x 10000020] 0 x 0000 0 xfeedbeef 0 x 00000000 l 6 Notice the byte ordering. This MIPS is “little-endian” (The least significant byte of a word or half-word has the lowest address) 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Assembler Syntax l Assembler DIRECTIVES (Keywords prefixed with a ‘. ’) – Control the placement and interpretation of bytes in memory . data <addr> Subsequent items are considered data. text <addr> Subsequent items are considered instructions. align N Skip to next address multiple of 2 N 7 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Assembler Syntax cont. – 8 Allocate Storage . byte b 1, b 2, …, bn Store a sequence of bytes (8 -bits). half h 1, h 2, …, hn Store a sequence of half-words (16 -bits). word w 1, w 2, …, wn Store a sequence of words (32 bits). ascii “string” Stores a sequence of ASCII encoded bytes. asciiz “string” Stores a zero-terminated string. space n Allocates n successive bytes 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Assembler Syntax cont. – Define scope . globl sym Declares symbol to be visible to other files. extern sym size Sets size of symbol defined in another file (Also makes it DIRECTLY addressable) 9 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Our Example: Variable Allocation l l 10 Two integer variables (by default 32 bits in MIPS). data. globl sum. globl i sum: . space 4 i: . space 4 “. data” assembler directive places the following words into the data segment “. globl” directives make the “sum” and “i” variables visible to all other assembly modules. “. space” directives allocate 4 bytes for each variable 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Actual “Code” l 11 Next we write ASSEMBLY code using the instruction Mnemonics. . text 0 x 80000080. globl main Main: add $8, $0 add $9, $0 Loop: addu $8, $9 addi $9, 1 slti $10, $9, 5 bne $10, $0, loop end: beq $0, end 19/02/2009 # sum = 0 # for (i = 0; . . . # sum = sum + i; # for (. . . ; i++ # for (. . . ; i<5; CA&O Lecture 05 by Engr. Umbreen Sabir

SPIM Simulation l l 12 Let’s tryout our Example We’ll use the PCSPIM (MIPS backwards) integrated ASSEMBLER, SIMULATOR, and DEBUGGER. 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Getting Started SPIMing l 13 The following hints will get you started with SPIM – By default a small fragment of code is loaded called the “kernel”. We will discuss the role of this code in detail, later in the course. Basically it’s primary job is to branch to the “main” label of your code. It does so in approximately 4 instructions. – We will use a raw version of the machine, w/o a kernel. – You can “single-step” the machine by pressing [F 10]. 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Getting Started SPIMing cont. – – 14 The results of the execution of each instruction are reflected in the register and memory location values. Illegal operations result in an “exception” which causes your code to automatically jump back the kernel. For our purposes now, this will be due to a bug in your program. Refer to the manual for more fancy usage, such as setting and executing to breakpoints 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

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“RAW” SPIM l 16 You’ll need to change the default settings as follows: 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

A Slightly More Challenging Program l l int i, sum; int a[5] = {7, 8, 9, 10, 8}; main() { sum = 0; for (i=0; i<5; i++) sum = sum + a[i]; } l Once more…let’s encode it in assembly l l l 17 Add 5 numbers from a list … sum = n 0 + n 1 + n 2 + n 3 + n 4 In “C”: 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Variable Allocation l l l l 18 l We cheated in our last example. Generally, variables will be allocated to memory locations, rather than registers. This time we add the contents of an array. data. extern sum 4. extern i 4. extern a 20 sum: . space 4 i: . space 4 a: . word 7, 8, 9, 10, 8 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Variable Allocation cont. l 19 “. word” allows us to initialize a list of sequential words in memory. The label represents the address of the first word in the list, or the name of the array 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

The New Code. text 0 x 80000080. globl main: sw $0, sum sw $0, i lw $9, i lw $8, sum loop: 20 # sum = 0; # for (i = 0; # allocate register for i # allocate register for sum 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

The New Code sll $10, $9, 2 lw $10, a($10) addu $8, $10 sw $8, sum addi $9, 1 sw $9, i slti $10, $9, 5 bne $10, $0, loop end: beq $0, end 21 # covert "i" to word offset # load a[i] # sum = sum + a[i]; # update variable in memory # for (. . . ; i++ # update memory # for (. . . ; i<5; 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

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A Coding Challenge What is the largest Fibonacci number less than 100? l Fibonacci numbers: Fi+1 = Fi + Fi-1 F 0 = 0 F 1 = 1 l 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, … l 23 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

A Coding Challenge l 24 In “C” int x, y; main() { x = 0; y = 1; while (y < 100) { int t = x; x = y; y = t + y; } } 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

Next Lecture and Reminders l Next lecture – MIPS ISA l 25 Assignment Due – 20 Feb 2009 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir

END OF LECTURE 5 26 19/02/2009 CA&O Lecture 05 by Engr. Umbreen Sabir