Difference between revisions of "CSC231 Final Exam 2010"
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| − | This final exam is take-home. It is open-books, open-notes, and open-Web. It is due a week after it is made available, at | + | This final exam is take-home. It is open-books, open-notes, and open-Web. It is due a week after it is made available, at 12:00 p.m. on Monday December 20, 2010. |
You cannot discuss the details of this exam with anyone except your instructor. The TAs are not allowed to help you out in any way. No question will be answered in person after 12:00 a.m. on 12/13/10. Instead, if you have questions regarding the exam, send them via email to thiebaut@cs.smith.edu, and the question and its answer will be broadcast back to the hole class via email. | You cannot discuss the details of this exam with anyone except your instructor. The TAs are not allowed to help you out in any way. No question will be answered in person after 12:00 a.m. on 12/13/10. Instead, if you have questions regarding the exam, send them via email to thiebaut@cs.smith.edu, and the question and its answer will be broadcast back to the hole class via email. | ||
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Make sure you reference all work/resources you use in your documentation. | Make sure you reference all work/resources you use in your documentation. | ||
| + | |||
| + | Files submitted past the deadline will not be graded. | ||
</tanbox> | </tanbox> | ||
| + | <br /> | ||
| + | <br /> | ||
| + | <br /> | ||
| + | <br /> | ||
| + | <br /> | ||
| + | __TOC__ | ||
| + | <br /> | ||
| + | <br /> | ||
| + | <br /> | ||
| + | <br /> | ||
| + | <onlydft> | ||
| + | =Problem #1: Recursive GCD (1 point)= | ||
| + | |||
| + | The following algorithm can be used to find the greatest common denominator of two integers, or ''GCD''. The GCD of two integers ''m'' and ''n'' is the largest integer that divides both of them with a remainder of 0. The GCD of 5 and 6 is 1. The GCD of 10 and 12 is 2. The GCD of 12 and 20 is 4. | ||
| + | |||
| + | int gcd(int m, int n) { | ||
| + | if ((m % n) == 0) | ||
| + | return n; | ||
| + | else | ||
| + | return gcd(n, m % n); | ||
| + | } | ||
| + | |||
| + | The % sign is the modulo operator. | ||
| + | |||
| + | Implement this algorithm in assembly using a recursive function called '''gcd'''. Make it compute and output the gcd of (5, 6), (10, 12), and (20, 30). | ||
| + | |||
| + | Call your program final1.asm and submit it as follows: | ||
| + | |||
| + | submit final final1.asm | ||
| + | |||
| + | |||
| − | + | =Problem #2: Debugging Utility Functions= | |
| − | =Problem # | + | ==Part 1 (1.5 points)== |
| − | ==Part 1== | ||
Write an assembly language program that does '''not''' use the driver.c or asm_io.asm files and that displays a 32-bit integer in binary, hexadecimal, and in decimal. When the number is displayed in decimal it is displayed as an ''unsigned'' '''int'''. For example, 0x80000000 should display as 2147483648. | Write an assembly language program that does '''not''' use the driver.c or asm_io.asm files and that displays a 32-bit integer in binary, hexadecimal, and in decimal. When the number is displayed in decimal it is displayed as an ''unsigned'' '''int'''. For example, 0x80000000 should display as 2147483648. | ||
| Line 40: | Line 72: | ||
mov ebx, table | mov ebx, table | ||
mov ecx, N | mov ecx, N | ||
| + | for: | ||
mov eax, [ebx] | mov eax, [ebx] | ||
| Line 61: | Line 94: | ||
:where nextLine is a function that brings the cursor to the next line. | :where nextLine is a function that brings the cursor to the next line. | ||
| − | ==Part 2== | + | ==Part 2 (1 point)== |
| − | Add a fourth function to your program that will display all the registers | + | Add a fourth function called '''printRegs''' to your program that will display all the registers in hex, in unsigned decimal, and in signed decimal formats. |
| + | |||
| + | Here is an example of the call and its output: | ||
| + | |||
| + | call printRegs | ||
| + | |||
| + | eax: 00000000 0 0 | ||
| + | ebx: 00001000 4096 4096 | ||
| + | ecx: 0fabbcde 262913246 262913246 | ||
| + | edx: 0000fffe 65534 65534 | ||
| + | esi: 00001132 4402 4402 | ||
| + | edi: ffffffff 4294967295 -1 | ||
| + | ebp: 0ffd1100 268243200 268243200 | ||
| + | esp: 0ffd1100 268243200 268243200 | ||
| + | |||
| + | Your function should display the value the registers hold before the call is made to printRegs. Your function should return to the calling program without having changed the value of any of the registers. | ||
| + | |||
| + | ==Part 3 (0.5 point)== | ||
| + | |||
| + | Add a new feature to the '''printRegs''' function so that it displays the flag bits. | ||
| + | |||
| + | The flag bits are located at specific positions in the flags register: | ||
| + | The EFLAGS register | ||
| + | |||
| + | Taken from http://www.eecg.toronto.edu/~amza/www.mindsec.com/files/x86regs.html: ''The EFLAGS register hold the state of the processor. It is modified by many intructions and is used for comparing some parameters, conditional loops and conditionnal jumps. Each bit holds the state of specific parameter of the last instruction. Here is a listing :'' | ||
| + | |||
| + | ''Bit Label Desciption | ||
| + | --------------------------- | ||
| + | 0 CF Carry flag | ||
| + | 2 PF Parity flag | ||
| + | 4 AF Auxiliary carry flag | ||
| + | 6 ZF Zero flag | ||
| + | 7 SF Sign flag | ||
| + | 8 TF Trap flag | ||
| + | 9 IF Interrupt enable flag | ||
| + | 10 DF Direction flag | ||
| + | 11 OF Overflow flag | ||
| + | 12-13 IOPL I/O Priviledge level | ||
| + | 14 NT Nested task flag | ||
| + | 16 RF Resume flag | ||
| + | 17 VM Virtual 8086 mode flag | ||
| + | 18 AC Alignment check flag (486+) | ||
| + | 19 VIF Virutal interrupt flag | ||
| + | 20 VIP Virtual interrupt pending flag | ||
| + | 21 ID ID flag'' | ||
| + | |||
| + | ''Those that are not listed are reserved by Intel.'' | ||
| + | |||
| + | You need to display only the Carry, Parity, Zero, and Sign bits. | ||
| + | |||
| + | Make sure that your function displays the status of the bits as they stand before the function is called, and make sure that whatever your function does, when it returns to the caller the status bits are in the same state they were before the function was called. | ||
| + | |||
| + | You will need the '''pushf''' and (maybe) the '''popf''' instructions. Pushf pushes a doubleword containing the flags register in the stack. Popf pops it back. | ||
| + | |||
| + | =Requirements= | ||
| + | |||
| + | * Submit only code that works | ||
| + | * Document your code well: | ||
| + | ** Functions should have headers | ||
| + | ** The main program should have a header. | ||
| + | |||
| + | =Submission= | ||
| + | Submit only one program for all parts. Call it final2.asm, and submit it as follows: | ||
| + | submit final final2.asm | ||
</onlydft> | </onlydft> | ||
Latest revision as of 08:38, 4 December 2014
This final exam is take-home. It is open-books, open-notes, and open-Web. It is due a week after it is made available, at 12:00 p.m. on Monday December 20, 2010.
You cannot discuss the details of this exam with anyone except your instructor. The TAs are not allowed to help you out in any way. No question will be answered in person after 12:00 a.m. on 12/13/10. Instead, if you have questions regarding the exam, send them via email to thiebaut@cs.smith.edu, and the question and its answer will be broadcast back to the hole class via email. The exam is given under the rules of the Smith College Honor Code.
Make sure you reference all work/resources you use in your documentation.
Files submitted past the deadline will not be graded.
