CSE 30 Final Exam Summer 2018

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CSE 30

Final Exam

Summer 2018

Problem 1 (9 pts): Number Systems a) (3 pts). Convert 0xFFCE (2’s complement format, 16-bit word) to the following. binary_____________________________________ (straight base conversion to binary) octal_______________________________________ (straight base conversion) decimal____________________________________ (convert to signed decimal) b) (3 pts). Convert 152 to the following (assume 16-bit word). Express answers in hexadecimal. sign-magnitude_______________________________________________ 1’s complement_______________________________________________ 2’s complement_______________________________________________ c) (3 pts). Convert -633 to the following (assume 16-bit word). Express answers in hexadecimal. sign-magnitude_______________________________________________ 1’s complement_______________________________________________ 2’s complement_______________________________________________

Problem 2 (4 pts): Rt-Lt Rule a) (2 pts). Using the C Rt-Lt Rule, define a variable named foo that is a pointer to a function which has a single parameter as a pointer to a struct fubar and returns a pointer to an array of 20 elements where each element  is of type int pointer.

b) (2 pts). Using the Rt-Lt Rule, write the C variable definition for the variable named foo that is a pointer to a 2D array with 10 rows and 20 columns where each element of the array is a pointer to a function that has no     parameter and returns an int.

Problem 3 (9 pts): Binary Addition/Condition Code Bits/Overflow Detection

Indicate what the condition code bits are when adding the following 8-bit 2’s complement numbers. Also provide the sum for each problem.

11010101

10001111

------------

N

Z

V

C

11111111

+         01111111

------------

N

Z

V

C

11110000

+         00001111

------------

N

Z

V

C

Problem 4 (6 pts): Bitwise Operations

What is the value (in hex) of r1 after each set of instructions. Assume that r0 and r1 start with the value in the comment section.  Make sure you write your answer as 32 bit words.

a). @ r0 = 0xDEADFACE

and r1, r0, #0xBABEFACE

Value in r1 at this point is 0x______________________________________________

b). @ r0 = 0XBEEFFACE

@ r1 = 0xDEEDBABE

eor r1, r0, r1

Value in r1 at this point is 0x______________________________________________

c). @r0 = 0x02468ACE

@r1 = 0x13579BDF

orr r1, r0, r1

Value in r1 at this point is 0x______________________________________________

d). @ r0 = 0xBAD5BABE

lsl r1, r0, #12

Value in r1 at this point is 0x______________________________________________

e). @ r0 = 0x87654321

asr r1, r0, #9

Value in r1 at this point is 0x______________________________________________

f). @ r0 = 0xBEEFBABE

lsr r1, r0, #4

Value in  r1 at this point is 0x______________________________________________

Problem 5 (27 pts): if statements with short-circuiting

Write the correct instructions to form the correct

sequence to translate the C code below into the

equivalent ARM assembly code. You should use the

approach we covered in lectures to translate. You

should assume that all local variables are mapped to

the stack. There are exactly the same number of lines

you should use. You should always assume the

independence of statements. For example, if a

variable needs to be modified, you should always

load from the memory no matter what you have

stored in your registers in previous statements.

// C code

/* Assume variables a and b have

been properly declared as ints. A

is the first declared local

variable, and b is the second

declared local variable.*/

if (a + b > 5 || b - 1 == 10){

b = a++ + 10;

}

else{

b = ++a + 5;

}

a++;

@ARM Assembly

@use r0 to manipulate a

@use r1 to manipulate b

@use r2 as temporary storage

@one instruction is given

ldr ____________________

ldr ____________________

add ____________________

____________________

____________________

ldr ____________________

____________________

____________________

____________________

@continue on the right

Problem 6 (20 pts): C runtime environment

Given the following program, reorder the output so that the address values that are printed are sorted from         smallest to largest if compiled and run on our pi-cluster. These lines print out the hex address of the different     parts of the program (not the values assigned) with the printf() format specifier %p (pointer). Basically, where do the different parts of a C program live in the run time environment? Write the correct letters in the answer    area.

#include <stdio.h>

int y;

int foo (int a, char b, short c)

{

int d;

printf("&a = %p\n", &a);          //A

printf("&b = %p\n", &b);          //B

printf("&c = %p\n", &c);          //C

printf("&d = %p\n", &d);          //D

}

int fubar (int x1, int x2, int x3, int x4,

int x5, int x6, int x7)

{

int tmp;

double r;

printf("&x1 = %p\n", &x1);

printf("&x2 = %p\n", &x2);

printf("&x3 = %p\n", &x3);

printf("&x4 = %p\n", &x4);

printf("&x5 = %p\n", &x5);

printf("&x6 = %p\n", &x6);

printf("&x7 = %p\n", &x7);

printf("&tmp = %p\n", &tmp);

printf("&r = %p\n", &r);

}

int main(int argc, char* argv[])

{

int m;

char n;

static short x = 10;

printf("main = %p\n", main);

printf("&x = %p\n", &x);

printf("&y = %p\n", &y);

printf("&argc = %p\n", &argc);

printf("&argv = %p\n", &argv);

printf("&m = %p\n", &m);

printf("&n = %p\n", &n);

foo(1,2,3);

fubar(x, y, 3, 4, 5, 6, 7);

}

Problem 7 (29 pts). Stack Frame


Given the C code below, first complete the memory stack. Clearly mark the location of fp and sp in the stack frame by writing sp and fp on the left side of the memory location it points to. Then fill in the blanks to         complete the translation from C to ARM assembly. Part B is on the next page.





2022-09-04