Difference between revisions of "CSC231 Final Exam 2010"
Line 104: | Line 104: | ||
ebx: 00001000 4096 4096 | ebx: 00001000 4096 4096 | ||
ecx: 0fabbcde 262913246 262913246 | ecx: 0fabbcde 262913246 262913246 | ||
− | + | edx: 0000fffe 65534 65534 | |
esi: 00001132 4402 4402 | esi: 00001132 4402 4402 | ||
edi: ffffffff 4294967295 -1 | edi: ffffffff 4294967295 -1 |
Revision as of 15:11, 16 December 2010
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.
Contents
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
Part 1 (1.5 points)
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.
Your program should contain three functions, one that receives the 32-bit integer in eax and displays it in binary. One that receives the 32-bit integer in eax and displays in hexadecimal. The third one will receive the 32-bit integer in eax and display it in decimal, without leading zeros.
Demonstrate the functionality of your program by making it display the following 32-bit integers:
1, 0xF, 0X10 0x12345678 0x7ffffff 0x89abcdef 0xffffffff
Here is what your main program should look like:
section .data
table dd 1, 15, 16
dd 0x12345678, 0x7ffffff, 0x89abcdef, 0xffffffff,
N equ ($-table)/4
section .text
start:
mov ebx, table
mov ecx, N
for:
mov eax, [ebx]
call displayBin
call nextLine
call displayHex
call nextLine
call displayUInt
call nextLine
call nextLine
add ebx, 4
loop for
mov eax, EXIT
mov ebx, 0
int 0x80
- where nextLine is a function that brings the cursor to the next line.
Part 2 (1 point)
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 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