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Chapter 5 LC3 Programming Solving Problems using a Computer Methodologies for creating computer programs that perform a desired function. Problem Solving • How do we figure out what to tell the computer to do? • Convert problem statement into algorithm, using stepwise refinement. • Convert algorithm into LC-3 machine instructions. Debugging • How do we figure out why it didn’t work? • Examining registers and memory, setting breakpoints, etc. Time spent on the first can reduce time spent on the second! 4-2 Stepwise Refinement Also known as systematic decomposition. Start with problem statement: “We wish to count the number of occurrences of a character in a file. The character in question is to be input from the keyboard; the result is to be displayed on the monitor.” Decompose task into a few simpler subtasks. Decompose each subtask into smaller subtasks, and these into even smaller subtasks, etc.... until you get to the machine instruction level. 4-3 Problem Statement Because problem statements are written in English, they are sometimes ambiguous and/or incomplete. • Where is “file” located? How big is it, or how do I know when I’ve reached the end? • How should final count be printed? A decimal number? • If the character is a letter, should I count both upper-case and lower-case occurrences? How do you resolve these issues? • Ask the person who wants the problem solved, or • Make a decision and document it. 4-4 Three Basic Constructs There are three basic ways to decompose a task: Task True False Test condition Subtask 1 Test condition False True Subtask 1 Subtask 2 Subtask 2 Sequential Subtask Conditional Iterative 4-5 Sequential Do Subtask 1 to completion, then do Subtask 2 to completion, etc. Get character input from keyboard Count and print the occurrences of a character in a file Examine file and count the number of characters that match Print number to the screen 4-6 Conditional If condition is true, do Subtask 1; else, do Subtask 2. True Test character. If match, increment counter. file char = input? False Count = Count + 1 4-7 Iterative Do Subtask over and over, as long as the test condition is true. Check each element of the file and count the characters that match. more chars to check? False True Check next char and count if matches. 4-8 Problem Solving Skills Learn to convert problem statement into step-by-step description of subtasks. • Like a puzzle, or a “word problem” from grammar school math. What is the starting state of the system? What is the desired ending state? How do we move from one state to another? • Recognize English words that correlate to three basic constructs: “do A then do B” sequential “if G, then do H” conditional “for each X, do Y” iterative “do Z until W” iterative 4-9 LC-3 Control Instructions How do we use LC-3 instructions to encode the three basic constructs? Sequential • Instructions naturally flow from one to the next, so no special instruction needed to go from one sequential subtask to the next. Conditional and Iterative • Create code that converts condition into N, Z, or P. Example: Condition: “Is R0 = R1?” Code: Subtract R1 from R0; if equal, Z bit will be set. • Then use BR instruction to transfer control to the proper subtask. 4-10 Code for Conditional Exact bits depend on condition being tested True Test Condition Subtask 1 PC offset to address C Instruction A False Generate Condition B 0000 ? C Subtask 1 Subtask 2 0000 111 D C Subtask 2 Next Subtask Unconditional branch to Next Subtask D Next Subtask Assuming all addresses are close enough that PC-relative branch can be used. PC offset to address D 4-11 Code for Iteration Exact bits depend on condition being tested Test Condition Instruction A False Generate Condition 0000 True ? C B Subtask Subtask 0000 111 C Next Subtask PC offset to address C Unconditional branch to retest condition Assuming all addresses are on the same page. A Next Subtask PC offset to address A 4-12 Example: Counting Characters START A START - Input a character. - Set up a pointer to the first location of the file that will be scanned. - Get the first character from the file. - Zero the register that holds the count. Input a character. Then scan a file, counting occurrences of that character. Finally, display on the monitor the number of occurrences of the character (up to 9). B STOP C Initial refinement: Big task into three sequential subtasks. Initialize: Put initial values into all locations that will be needed to carry out this task. Scan the file, location by location, incrementing the counter if the character matches. Display the count on the monitor. STOP 4-13 Refining B B Yes B Scan the file, location by location, incrementing the counter if the character matches. B1 Done? No Test character. If a match, increment counter. Get next character. Refining B into iterative construct. 4-14 Refining B1 Yes B Yes B1 Done? No Test character. If a match, increment counter. Get next character. Done? No B1 B2 Test character. If matches, increment counter. B3 Get next character. Refining B1 into sequential subtasks. 4-15 Refining B2 and B3 Yes Done? No B2 Yes Yes Done? No R1 = R0? No R2 = R2 + 1 B1 B2 Test character. If matches, increment counter. B3 B3 Get next character. R3 = R3 + 1 R1 = M[R3] Conditional (B2) and sequential (B3). Use of LC-2 registers and instructions. 4-16 The Last Step: LC-3 Instructions Use comments to separate into modules and to document your code. Yes Done? No B2 Yes R1 = R0? R2 = R2 + 1 B3 R3 = R3 + 1 No ; Look at each char in file. 0001100001111100 ; is R1 = EOT? 0000010xxxxxxxxx ; if so, exit loop ; Check for match with R0. 1001001001111111 ; R1 = -char 0001001001100001 0001001000000001 ; R1 = R0 – char 0000101xxxxxxxxx ; no match, skip incr 0001010010100001 ; R2 = R2 + 1 ; Incr file ptr and get next char 0001011011100001 ; R3 = R3 + 1 0110001011000000 ; R1 = M[R3] R1 = M[R3] Don’t know PCoffset bits until all the code is done 4-17 Debugging You’ve written your program and it doesn’t work. Now what? What do you do when you’re lost in a city? Drive around randomly and hope you find it? Return to a known point and look at a map? In debugging, the equivalent to looking at a map is tracing your program. • Examine the sequence of instructions being executed. • Keep track of results being produced. • Compare result from each instruction to the expected result. 4-18 Debugging Operations Any debugging environment should provide means to: 1. 2. 3. 4. Display values in memory and registers. Deposit values in memory and registers. Execute instruction sequence in a program. Stop execution when desired. Different programming levels offer different tools. • • High-level languages (C, Java, ...) usually have source-code debugging tools. For debugging at the machine instruction level: simulators operating system “monitor” tools in-circuit emulators (ICE) – plug-in hardware replacements that give instruction-level control 4-19 LC-3 Simulator execute instruction sequences stop execution, set breakpoints set/display registers and memory 4-20 Types of Errors Syntax Errors • You made a typing error that resulted in an illegal operation. • Not usually an issue with machine language, because almost any bit pattern corresponds to some legal instruction. • In high-level languages, these are often caught during the translation from language to machine code. Logic Errors • Your program is legal, but wrong, so the results don’t match the problem statement. • Trace the program to see what’s really happening and determine how to get the proper behavior. Data Errors • Input data is different than what you expected. • Test the program with a wide variety of inputs. 4-21 Tracing the Program Execute the program one piece at a time, examining register and memory to see results at each step. Single-Stepping • Execute one instruction at a time. • Tedious, but useful to help you verify each step of your program. Breakpoints • Tell the simulator to stop executing when it reaches a specific instruction. • Check overall results at specific points in the program. Lets you quickly execute sequences to get a high-level overview of the execution behavior. Quickly execute sequences that your believe are correct. Watchpoints • Tell the simulator to stop when a register or memory location changes or when it equals a specific value. • Useful when you don’t know where or when a value is changed. 4-22 Example 1: Multiply This program is supposed to multiply the two unsigned integers in R4 and R5. clear R2 add R4 to R2 decrement R5 No x3200 x3201 x3202 x3203 x3204 0101010010100000 0001010010000100 0001101101111111 0000011111111101 1111000000100101 R5 = 0? Yes HALT Set R4 = 10, R5 =3. Run program. Result: R2 = 40, not 30. 4-23 Debugging the Multiply Program Single-stepping PC PC and registers at the beginning of each instruction R2 R4 R5 Breakpoint at branch (x3203) x3200 -- 10 3 x3201 0 10 3 x3202 10 10 3 PC x3203 10 10 2 x3203 10 10 2 x3201 10 10 2 x3203 20 10 1 x3202 20 10 2 x3203 30 10 0 x3203 20 10 1 x3203 40 10 -1 x3201 20 10 1 40 10 -1 x3202 30 10 1 x3203 30 10 0 x3201 30 10 0 x3202 40 10 0 x3203 40 10 -1 x3204 40 10 -1 40 10 -1 R2 R4 R5 Should stop looping here! Executing loop one time too many. Branch at x3203 should be based on Z bit only, not Z and P. 4-24 Example 2: Summing an Array of Numbers This program is supposed to sum the numbers stored in 10 locations beginning with x3100, leaving the result in R1. R1 = 0 R4 = 10 R2 = x3100 R1 = R1 + M[R2] R2 = R2 + 1 R4 = R4 - 1 No R4 = 0? Yes HALT x3000 x3001 x3002 x3003 x3004 x3005 x3006 x3007 x3008 x3009 0101001001100000 0101100100100000 0001100100101010 0010010011111100 0110011010000000 0001010010100001 0001001001000011 0001100100111111 0000001111111011 1111000000100101 4-25 Debugging the Summing Program Running the the data below yields R1 = x0024, but the sum should be x8135. What happened? Address Contents x3100 x3107 x3101 x2819 x3000 -- -- -- x3102 x0110 x3001 0 -- -- x3103 x0310 x3002 0 -- 0 x3003 0 -- 10 x3104 x0110 x3004 0 x3107 10 x3105 x1110 x3106 x11B1 x3107 x0019 x3108 x0007 x3109 x0004 Start single-stepping program... PC R1 R2 R4 Should be x3100! Loading contents of M[x3100], not address. Change opcode of x3003 from 0010 (LD) to 1110 (LEA). 4-26 Example 3: Looking for a 5 This program is supposed to set R0=1 if there’s a 5 in one ten memory locations, starting at x3100. Else, it should set R0 to 0. R0 = 1, R1 = -5, R3 = 10 R4 = x3100, R2 = M[R4] R2 = 5? No No R3 = 0? R4 = R4 + 1 R3 = R3-1 R2 = M[R4] Yes R0 = 0 HALT Yes x3000 x3001 x3002 x3003 x3004 x3005 x3006 x3007 x3008 x3009 x300A x300B x300C x300D x300E x300F x3010 0101000000100000 0001000000100001 0101001001100000 0001001001111011 0101011011100000 0001011011101010 0010100000001001 0110010100000000 0001010010000001 0000010000000101 0001100100100001 0001011011111111 0110010100000000 0000001111111010 0101000000100000 1111000000100101 0011000100000000 4-27 Debugging the Fives Program Running the program with a 5 in location x3108 results in R0 = 0, not R0 = 1. What happened? Perhaps we didn’t look at all the data? Put a breakpoint at x300D to see how many times we branch back. Address Contents x3100 9 x3101 7 x3102 32 x300D 1 7 9 x3101 x3103 0 x300D 1 32 8 x3102 x3104 -8 x300D 1 0 7 x3103 0 0 7 x3103 x3105 19 x3106 6 x3107 13 x3108 5 x3109 61 PC R0 R2 R3 R4 Didn’t branch back, even though R3 > 0? Branch uses condition code set by loading R2 with M[R4], not by decrementing R3. Swap x300B and x300C, or remove x300C and branch back to x3007. 4-28 Example 4: Finding First 1 in a Word This program is supposed to return (in R1) the bit position of the first 1 in a word. The address of the word is in location x3009 (just past the end of the program). If there are no ones, R1 should be set to –1. R1 = 15 R2 = data R2[15] = 1? No decrement R1 shift R2 left one bit No R2[15] = 1? Yes HALT Yes x3000 x3001 x3002 x3003 x3004 x3005 x3006 x3007 x3008 x3009 0101001001100000 0001001001101111 1010010000000110 0000100000000100 0001001001111111 0001010010000010 0000100000000001 0000111111111100 1111000000100101 0011000100000000 4-29 Debugging the First-One Program Program works most of the time, but if data is zero, it never seems to HALT. Breakpoint at backwards branch (x3007) PC R1 PC R1 x3007 14 x3007 4 x3007 13 x3007 3 x3007 12 x3007 2 x3007 11 x3007 1 x3007 10 x3007 0 x3007 9 x3007 -1 x3007 8 x3007 -2 x3007 7 x3007 -3 x3007 6 x3007 -4 x3007 5 x3007 -5 If no ones, then branch to HALT never occurs! This is called an “infinite loop.” Must change algorithm to either (a) check for special case (R2=0), or (b) exit loop if R1 < 0. 4-30 Debugging: Lessons Learned Trace program to see what’s going on. • Breakpoints, single-stepping When tracing, make sure to notice what’s really happening, not what you think should happen. • In summing program, it would be easy to not notice that address x3107 was loaded instead of x3100. Test your program using a variety of input data. • In Examples 3 and 4, the program works for many data sets. • Be sure to test extreme cases (all ones, no ones, ...). 4-31 2. Assembly Language Human-Readable Machine Language Computers like ones and zeros… 0001110010000110 Humans like symbols… ADD R6,R2,R6 ; increment index reg. Assembler is a program that turns symbols into machine instructions. • ISA-specific: close correspondence between symbols and instruction set mnemonics for opcodes labels for memory locations • additional operations for allocating storage and initializing data 4-33 An Assembly Language Program ; ; Program to multiply a number by the constant 6 ; .ORIG x3050 LD R1, SIX LD R2, NUMBER AND R3, R3, #0 ; Clear R3. It will ; contain the product. ; The inner loop ; AGAIN ADD R3, R3, R2 ADD R1, R1, #-1 ; R1 keeps track of BRp AGAIN ; the iteration. ; HALT ; NUMBER .BLKW 1 SIX .FILL x0006 ; .END 4-34 LC-3 Assembly Language Syntax Each line of a program is one of the following: • an instruction • an assember directive (or pseudo-op) • a comment Whitespace (between symbols) and case are ignored. Comments (beginning with “;”) are also ignored. An instruction has the following format: LABEL OPCODE OPERANDS ; COMMENTS optional mandatory 4-35 Opcodes and Operands Opcodes • reserved symbols that correspond to LC-3 instructions • listed in Appendix A ex: ADD, AND, LD, LDR, … Operands • • • • • registers -- specified by Rn, where n is the register number numbers -- indicated by # (decimal) or x (hex) label -- symbolic name of memory location separated by comma number, order, and type correspond to instruction format ex: ADD R1,R1,R3 ADD R1,R1,#3 LD R6,NUMBER BRz LOOP 4-36 Labels and Comments Label • placed at the beginning of the line • assigns a symbolic name to the address corresponding to line ex: LOOP ADD R1,R1,#-1 BRp LOOP Comment • • • • anything after a semicolon is a comment ignored by assembler used by humans to document/understand programs tips for useful comments: avoid restating the obvious, as “decrement R1” provide additional insight, as in “accumulate product in R6” use comments to separate pieces of program 4-37 Assembler Directives Pseudo-operations • do not refer to operations executed by program • used by assembler • look like instruction, but “opcode” starts with dot Opcode Operand Meaning .ORIG address starting address of program .END end of program .BLKW n allocate n words of storage .FILL n allocate one word, initialize with value n .STRINGZ n-character string allocate n+1 locations, initialize w/characters and null terminator 4-38 Trap Codes LC-3 assembler provides “pseudo-instructions” for each trap code, so you don’t have to remember them. Code Equivalent Description HALT TRAP x25 Halt execution and print message to console. IN TRAP x23 Print prompt on console, read (and echo) one character from keybd. Character stored in R0[7:0]. OUT TRAP x21 Write one character (in R0[7:0]) to console. GETC TRAP x20 Read one character from keyboard. Character stored in R0[7:0]. PUTS TRAP x22 Write null-terminated string to console. Address of string is in R0. 4-39 Style Guidelines Use the following style guidelines to improve the readability and understandability of your programs: 1. Provide a program header, with author’s name, date, etc., and purpose of program. 2. Start labels, opcode, operands, and comments in same column for each line. (Unless entire line is a comment.) 3. Use comments to explain what each register does. 4. Give explanatory comment for most instructions. 5. Use meaningful symbolic names. • Mixed upper and lower case for readability. • ASCIItoBinary, InputRoutine, SaveR1 6. Provide comments between program sections. 7. Each line must fit on the page -- no wraparound or truncations. • Long statements split in aesthetically pleasing manner. 4-40 Sample Program Count the occurrences of a character in a file. Remember this? Count = 0 (R2 = 0) Done? YES (R1 ?= EOT) Ptr = 1st file character Convert count to ASCII character (R0 = x30, R0 = R2 + R0) NO (R3 = M[x3012]) Print count YES Match? NO (TRAP x21) (R1 ?= R0) Input char from keybd (TRAP x23) HALT Incr Count Load char from file (TRAP x25) (R2 = R2 + 1) (R1 = M[R3]) Load next char from file (R3 = R3 + 1, R1 = M[R3]) 4-41 Char Count in Assembly Language (1 of 3) ; ; ; ; ; ; ; ; ; Program to count occurrences of a character in a file. Character to be input from the keyboard. Result to be displayed on the monitor. Program only works if no more than 9 occurrences are found. Initialization .ORIG AND LD GETC LDR x3000 R2, R2, #0 R3, PTR R1, R3, #0 ; ; ; ; R2 R3 R0 R1 is counter, initially 0 is pointer to characters gets character input gets first character ; ; Test character for end of file ; TEST ADD R4, R1, #-4 ; Test for EOT (ASCII x04) BRz OUTPUT ; If done, prepare the output 4-42 Char Count in Assembly Language (2 of 3) ; ; Test character for match. If a match, increment count. ; NOT R1, R1 ADD R1, R1, R0 ; If match, R1 = xFFFF NOT R1, R1 ; If match, R1 = x0000 BRnp GETCHAR ; If no match, do not increment ADD R2, R2, #1 ; ; Get next character from file. ; GETCHAR ADD R3, R3, #1 ; Point to next character. LDR R1, R3, #0 ; R1 gets next char to test BRnzp TEST ; ; Output the count. ; OUTPUT LD R0, ASCII ; Load the ASCII template ADD R0, R0, R2 ; Covert binary count to ASCII OUT ; ASCII code in R0 is displayed. HALT ; Halt machine 4-43 Char Count in Assembly Language (3 of 3) ; ; Storage for pointer and ASCII template ; ASCII .FILL x0030 PTR .FILL x4000 .END 4-44 Assembly Process Convert assembly language file (.asm) into an executable file (.obj) for the LC-3 simulator. First Pass: • scan program file • find all labels and calculate the corresponding addresses; this is called the symbol table Second Pass: • convert instructions to machine language, using information from symbol table 4-45 First Pass: Constructing the Symbol Table 1. Find the .ORIG statement, which tells us the address of the first instruction. • Initialize location counter (LC), which keeps track of the current instruction. 2. For each non-empty line in the program: a) If line contains a label, add label and LC to symbol table. b) Increment LC. – NOTE: If statement is .BLKW or .STRINGZ, increment LC by the number of words allocated. 3. Stop when .END statement is reached. NOTE: A line that contains only a comment is considered an empty line. 4-46 Second Pass: Generating Machine Language For each executable assembly language statement, generate the corresponding machine language instruction. • If operand is a label, look up the address from the symbol table. Potential problems: • Improper number or type of arguments ex: NOT R1,#7 ADD R1,R2 ADD R3,R3,NUMBER • Immediate argument too large ex: ADD R1,R2,#1023 • Address (associated with label) more than 256 from instruction can’t use PC-relative addressing mode 4-47 Practice Using the symbol table constructed earlier, translate these statements into LC-3 machine language. Statement LD R3,PTR ADD R4,R1,#-4 LDR R1,R3,#0 Machine Language BRnp GETCHAR 4-48 LC-3 Assembler Using “assemble” (Unix) or LC3Edit (Windows), generates several different output files. This one gets loaded into the simulator. 4-49 Object File Format LC-3 object file contains • Starting address (location where program must be loaded), followed by… • Machine instructions Example • Beginning of “count character” object file looks like this: 0011000000000000 0101010010100000 0010011000010001 1111000000100011 . . . .ORIG x3000 AND R2, R2, #0 LD R3, PTR TRAP x23 4-50 Multiple Object Files An object file is not necessarily a complete program. • system-provided library routines • code blocks written by multiple developers For LC-3 simulator, can load multiple object files into memory, then start executing at a desired address. • system routines, such as keyboard input, are loaded automatically loaded into “system memory,” below x3000 user code should be loaded between x3000 and xFDFF • each object file includes a starting address • be careful not to load overlapping object files 4-51 Linking and Loading Loading is the process of copying an executable image into memory. • more sophisticated loaders are able to relocate images to fit into available memory • must readjust branch targets, load/store addresses Linking is the process of resolving symbols between independent object files. • suppose we define a symbol in one module, and want to use it in another • some notation, such as .EXTERNAL, is used to tell assembler that a symbol is defined in another module • linker will search symbol tables of other modules to resolve symbols and complete code generation before loading 4-52