Slide hệ thống máy tính và ngôn ngữ lập trình c chương 5lc3 programming

Solving Problems using a Computer Methodologies for creating computer programs that perform a desired function Chapter Problem Solving • How we figure out what to tell the computer to do? • Convert problem statement into algorithm, algorithm using stepwise refinement • Convert algorithm into LC-3 machine instructions LC3 Programming Debugging • How 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 Problem Statement Also known as systematic decomposition Because problem statements are written in English, they are sometimes ambiguous and/or incomplete • Where is “file” located? How big is it, or how 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 d lower-case l occurrences? ? 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 How you resolve these issues? • Ask the person who wants the problem solved, or • Make a decision and document it Decompose each subtask into smaller subtasks, and these into even smaller subtasks, etc until you get to the machine instruction level 4-3 4-4 CuuDuongThanCong.com https://fb.com/tailieudientucntt Three Basic Constructs Sequential There are three basic ways to decompose a task: Do Subtask to completion, then Subtask to completion, etc Task Get character input from keyboard True False Test condition Subtask Test condition Count and print the occurrences of a character in a file False Examine file and count the number of characters that match True Subtask Subtask Subtask Sequential Subtask Conditional Iterative Print number to the screen 4-5 4-6 Conditional Iterative If condition is true, Subtask 1; else, Subtask Do Subtask over and over, as long as the test condition is true True Test character If match, increment counter file char = input? ? False Check each element of the file and count the characters that match Count = Count + more chars to check? False True Check next char and count if matches 4-7 4-8 CuuDuongThanCong.com https://fb.com/tailieudientucntt Problem Solving Skills LC-3 Control Instructions Learn to convert problem statement into step-by-step description of subtasks How we use LC-3 instructions to encode the three basic constructs? • Like a puzzle, or a “word problem” from grammar school math ¾What is the starting state of the system? ¾What is the desired ending state? ¾How we move from one state to another? • Recognize English words that correlate to three basic constructs: ¾“do A then B” ⇒ sequential ¾“if G, then H” ⇒ conditional ¾“for each X, Y” ⇒ iterative ¾“do Z until W” ⇒ iterative 4-9 Code for Conditional Test Condition • Instructions naturally flow from one to the next, so no special p instruction needed to go g 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 Iteration Exact bits depend on condition being tested True Sequential PC offset to address C Instruction A False Exact bits depend on condition being tested Generate Condition B 0000 ? Test Condition C Instruction A False Generate Condition 0000 Subtask Subtask Subtask 0000 111 True 0000 111 Subtask Unconditional branch to Next Subtask D Next Subtask Assuming all addresses are close enough that PC-relative branch can be used C PC offset to address D 4-11 Next Subtask C Subtask C Next Subtask ? B Subtask D 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 CuuDuongThanCong.com https://fb.com/tailieudientucntt Example: Counting Characters Refining B START A START Initialize: Put initial values into all locations that will be needed to carry out this task B - 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 Yes B Scan the file, location by l location, ti incrementing i ti the th counter if the character matches Done? No B1 Test character If a match, increment counter Get next character Scan the file, location by location, incrementing the counter if the character matches Display the count on the monitor Initial refinement: Big task into three sequential subtasks Refining B into iterative construct STOP 4-13 Refining B1 4-14 Refining B2 and B3 Yes Done? No B2 Yes B Done? Yes Yes B1 Done? No Test character If a match, increment counter Get next character No B1 Yes Done? N No No R2 = R2 + B1 B2 Test character If matches, increment counter R1 = R0? B2 Test character If matches, increment counter B3 B3 Get next character B3 Get next character R3 = R3 + R1 = M[R3] Refining B1 into sequential subtasks 4-15 Conditional (B2) and sequential (B3) Use of LC-2 registers and instructions 4-16 CuuDuongThanCong.com https://fb.com/tailieudientucntt The Last Step: LC-3 Instructions Debugging Use comments to separate into modules and to document your code You’ve written your program and it doesn’t work Now what? Yes Done? No B2 Yes R1 = R0? No R2 = R2 + B3 R3 = R3 + ; Look at each char in file 0001100001111100 ; is R1 = EOT? 0000010xxxxxxxxx ; if so, exit loop ; Check for match with R0 R0 1001001001111111 ; R1 = -char 0001001001100001 0001001000000001 ; R1 = R0 – char 0000101xxxxxxxxx ; no match, skip incr 0001010010100001 ; R2 = R2 + ; Incr file ptr and get next char 0001011011100001 ; R3 = R3 + 0110001011000000 ; R1 = M[R3] R1 = M[R3] Don’t know PCoffset bits until all the code is done What you when you’re lost in a city? Drive around randomly and hope you find it? 3Return 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 4-17 Debugging Operations 4-18 LC-3 Simulator Any debugging environment should provide means to: execute instruction sequences 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 • • set/display registers and memory High-level languages (C, Java, ) usually have source-code debugging tools For debugging at the machine instruction level: ¾ simulators ¾ operating system “monitor” tools ¾ in-circuit emulators (ICE) – plug-in hardware replacements that give instruction-level control stop execution, set breakpoints 4-19 4-20 CuuDuongThanCong.com https://fb.com/tailieudientucntt Types of Errors Tracing the Program Syntax Errors Execute the program one piece at a time, examining register and memory to see results at each step Single-Stepping • 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 code • 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 ¾ Lets you quickly execute sequences to get a high-level overview of the execution behavior ¾ Quickly execute sequences that your believe are correct 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 Watchpoints Data Errors • Input data is different than what you expected • Test the program with a wide variety of inputs • 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 4-21 Example 1: Multiply 4-22 Debugging the Multiply Program Single-stepping 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 PC and registers at the beginning of each instruction 0101010010100000 0001010010000100 0001101101111111 0000011111111101 1111000000100101 R5 = 0? Yes HALT PC Set R4 = 10, R5 =3 Run program Result: R2 = 40, not 30 R2 R4 R5 Breakpoint at branch (x3203) x3200 10 x3201 10 x3202 10 10 PC x3203 10 10 x3203 10 10 x3201 10 10 x3203 20 10 x3202 3202 20 10 x3203 3203 30 10 x3203 20 10 x3203 40 10 -1 x3201 20 10 40 10 -1 x3202 30 10 x3203 30 10 x3201 30 10 x3202 40 10 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 4-23 4-24 CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 2: Summing an Array of Numbers Debugging the Summing Program This program is supposed to sum the numbers stored in 10 locations beginning with x3100, leaving the result in R1 Running the the data below yields R1 = x0024, but the sum should be x8135 What happened? R1 = R4 = 10 R2 = x3100 x3000 x3001 x3002 x3003 x3004 x3005 x3006 x3007 x3008 x3009 R1 = R1 + M[R2] R2 = R2 + R4 = R4 - No R4 = 0? Yes 0101001001100000 0101100100100000 0001100100101010 0010010011111100 0110011010000000 0001010010100001 0001001001000011 0001100100111111 0000001111111011 1111000000100101 Example 3: Looking for a This program is supposed to set R0=1 if there’s a in one ten memory locations, starting at x3100 Else, it should set R0 to R0 = 1, R1 = -5, R3 = 10 R4 = x3100,, R2 = M[R4] [ ] R2 = 5? No R3 = 0? R4 = R4 + R3 = R3-1 R2 = M[R4] Yes R0 = Contents x3100 x3107 x3101 x2819 x3000 x3102 3102 x0110 0110 x3001 3001 x3103 x0310 x3002 x3003 10 x3104 x0110 x3004 x3107 10 x3105 x1110 x3106 x11B1 x3107 x0019 x3108 x0007 x3109 x0004 4-25 HALT No Address HALT Yes 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 Debugging the Fives Program x3000 x3001 x3002 x3003 x3004 x3005 x3006 x3007 3007 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 Running the program with a in location x3108 results in R0 = 0, not R0 = What happened? Address Contents x3100 Perhaps we didn’t look at all the data? Put a breakpoint at x300D to see how many times we branch back x3101 x3102 3102 32 x300D x3101 x3103 x300D 32 x3102 x3104 -8 x300D x3103 x3105 19 0 x3103 x3106 x3107 13 x3108 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 CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4: Finding First in a Word Debugging the First-One Program This program is supposed to return (in R1) the bit position of the first 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 Program works most of the time, but if data is zero, it never seems to HALT R2[15] = 1? Yes No decrement R1 shift R2 left one bit No Breakpoint at backwards branch (x3007) PC R1 = 15 R2 = data R2[15] = 1? Yes x3000 x3001 x3002 x3003 x3004 x3005 x3006 x3007 x3008 x3009 0101001001100000 0001001001101111 1010010000000110 0000100000000100 0001001001111111 0001010010000010 0000100000000001 0000111111111100 1111000000100101 0011000100000000 HALT R1 PC R1 x3007 14 x3007 x3007 13 x3007 3007 x3007 12 x3007 x3007 11 x3007 x3007 10 x3007 x3007 x3007 -1 x3007 x3007 -2 x3007 x3007 -3 x3007 x3007 -4 x3007 x3007 -5 If no ones, then th branch b h to t 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 < 4-29 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 Assembly Language • In summing program, 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 and 4, the program works for many data sets • Be sure to test extreme cases (all ones, no ones, ) 4-31 CuuDuongThanCong.com https://fb.com/tailieudientucntt Human-Readable Machine Language An Assembly Language Program Computers like ones and zeros… 0001110010000110 Humans like symbols… ; ; Program to multiply a number by the constant ; 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 SIX FILL x0006 ; END 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 ¾mnemonics for opcodes ¾labels for memory locations • additional operations for allocating storage and initializing data 4-33 4-34 LC-3 Assembly Language Syntax Opcodes and Operands Each line of a program is one of the following: Opcodes • reserved symbols that correspond to LC-3 instructions ã listed in Appendix A ắex: ADD, AND, LD, LDR, … • an instruction • an assember directive (or pseudo-op) • a comment Operands 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 registers specified by Rn, where n is the register number numbers b indicated i di d by b # (decimal) (d i l) or x (hex) (h ) label symbolic name of memory location separated by comma number, order, and type correspond to instruction format ¾ex: ADD R1,R1,R3 ADD R1,R1,#3 LD R6,NUMBER BRz LOOP 4-35 4-36 CuuDuongThanCong.com https://fb.com/tailieudientucntt Labels and Comments Assembler Directives Label Pseudo-operations • placed at the beginning of the line • assigns a symbolic name to the address corresponding to line ¾ex: LOOP ADD R1,R1,#-1 BRp LOOP C Comment t • • • • anything after a semicolon is a comment ignored by assembler used by humans to document/understand programs tips for useful comments: ¾avoid restating the obvious, as “decrement R1” ¾provide additional insight, as in “accumulate product in R6” ¾use comments to separate pieces of program • 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 off program end of program END 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 4-37 4-38 Trap Codes Style Guidelines LC-3 assembler provides “pseudo-instructions” for each trap code, so you don’t have to remember them Use the following style guidelines to improve the readability and understandability of your programs: Code Equivalent Description HALT TRAP x25 Halt execution and print message to console IN TRAP x23 Print prompt on console, 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 Provide a program header, with author’s name, date, etc., and purpose of program Start labels, opcode, operands, and comments in same column for each line (Unless entire line is a comment.) Use comments to explain what each register does Give explanatory comment for most instructions Use meaningful symbolic names • Mixed upper and lower case for readability • ASCIItoBinary, InputRoutine, SaveR1 Provide comments between program sections Each line must fit on the page no wraparound or truncations • Long statements split in aesthetically pleasing manner 4-39 4-40 10 CuuDuongThanCong.com https://fb.com/tailieudientucntt Sample Program Char Count in Assembly Language (1 of 3) Count the occurrences of a character in a file ; ; ; ; ; ; ; ; ; Remember this? Count = (R2 = 0) Done? YES (R1 ?= EOT) Ptr = 1st file character Convert count to ASCII character (R0 = x30, R0 = R2 + R0) Print count Match? NO (TRAP x21) (TRAP x23) HALT Incr Count Load char from file x3000 R2 R2 R2, R2, #0 R3, PTR ; ; ; ; R2 R3 R0 R1 i is counter, t i initially iti ll is pointer to characters gets character input gets first character R1, R3, #0 ; ; Test character for end of file ; TEST ADD R4, R1, #-4 ; Test for EOT (ASCII x04) BRz OUTPUT ; If done, prepare the output (R1 ?= R0) Input char from keybd Initialization ORIG AND LD GETC LDR NO (R3 = M[x3012]) YES 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 occurrences are found (TRAP x25) (R2 = R2 + 1) (R1 = M[R3]) Load next char from file (R3 = R3 + 1, R1 = M[R3]) 4-41 4-42 Char Count in Assembly Language (2 of 3) Char Count in Assembly Language (3 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, 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 ; ; Storage for pointer and ASCII template ; ASCII FILL x0030 PTR FILL x4000 END 4-43 4-44 11 CuuDuongThanCong.com https://fb.com/tailieudientucntt Assembly Process First Pass: Constructing the Symbol Table Convert assembly language file (.asm) into an executable file (.obj) for the LC-3 simulator Find the ORIG statement, which tells us the address of the first instruction • Initialize location counter (LC), which keeps track of the current instruction For each non-empty line in the program: a) If line contains a label, label add label and LC to symbol table table b) Increment LC – NOTE: If statement is BLKW or STRINGZ, increment LC by the number of words allocated First Pass: • scan program file • find all labels and calculate the corresponding addresses; this is called the symbol table Stop when END statement is reached Second Pass: • convert instructions to machine language, using information from symbol table NOTE: A line that contains only a comment is considered an empty line 4-45 4-46 Second Pass: Generating Machine Language Practice For each executable assembly language statement, generate the corresponding machine language instruction Using the symbol table constructed earlier, translate these statements into LC-3 machine language • If operand is a label, look up the address from the symbol table Statement Potential problems: • Improper number or type of arguments ắex: NOT R1,#7 ADD R1,R2 ADD R3,R3,NUMBER ã Immediate argument too large ắex: ADD R1,R2,#1023 ã Address (associated with label) more than 256 from instruction ¾can’t use PC-relative addressing mode LD R3,PTR ADD R4,R1,#-4 LDR R1,R3,#0 Machine Language BRnp GETCHAR 4-47 4-48 12 CuuDuongThanCong.com https://fb.com/tailieudientucntt LC-3 Assembler Object File Format Using “assemble” (Unix) or LC3Edit (Windows), generates several different output files LC-3 object file contains This one gets loaded into the simulator • Starting address (location where program must be loaded), followed by… • Machine instructions Example p • Beginning of “count character” object file looks like this: 4-49 0011000000000000 0101010010100000 0010011000010001 1111000000100011 ORIG x3000 AND R2, R2, #0 LD R3, PTR TRAP x23 4-50 Multiple Object Files Linking and Loading An object file is not necessarily a complete program Loading is the process of copying an executable image into memory • system-provided library routines • code blocks written by multiple developers • more sophisticated loaders are able to relocate images to fit into available memory • must readjust branch targets, load/store addresses For LC-3 simulator, can load multiple object files into memory, then start executing at a desired address Linking is the process of resolving symbols between independent object files • system routines, such as keyboard input, are loaded automatically ¾loaded into “system memory,” below x3000 ¾user code should be loaded between x3000 and xFDFF • each object file includes a starting address • be careful not to load overlapping 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 4-51 4-52 13 CuuDuongThanCong.com https://fb.com/tailieudientucntt ... file char = input? ? False Check each element of the file and count the characters that match Count = Count + more chars to check? False True Check next char and count if matches 4-7 4-8 CuuDuongThanCong.com... 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 Yes B Scan the... location by l location, ti incrementing i ti the th counter if the character matches Done? No B1 Test character If a match, increment counter Get next character Scan the file, location by location,