Sample Midterm Solution

Cs 216, Midterm exam, (from 2005) But expect some C

Actual midterm solution

Open book You may consult your textbook, my printed class notes, and one page of information you have prepared. You may also use a printed dictionary. You may not use a calculator or any other electronic device. Do not copy any wrong answers!

Convert this signed hexadecimal number to decimal, show your work: 0xFFFFFE60

It is negative, so

F F F F E 6 0
0 0 0 0 1 9 F  - one's complement
0 0 0 0 1 A 0  - two's complement
	1 	*16²	=  256
	 10	*16	=  160 
	    0
			   416,		so the number is -416

(Corrected. Thank you, Simon.)

Convert the decimal number 444 to binary (or hexadecimal), show your work:

444/16	= 27  R 12 = C
 27/16  =  1  R 11 = B
  1/16	=  0  R  1 = 1   The hex number is  0x1BC or 1 1011 1100

Given that $a0 contains an integer, If it is negative, put its absolute value in $v0, and put 0 (zero) in $v1. However, if $a0 is positive, put the remainder of dividing by 128 in $v1, and the square of the quotient in $v0. Note that $v0 and $v1 must still be as specified when the end of your section of code is finished. (You are not to write a whole program, just the instructions to do as asked.)

Good grief, well anyway this looks like an IF - then - else situation
```
bgez	$a0, positive
	abs	$v0, $a0	#absolute value of negative number
	li	$v1, 0		# 0 as directed
	j	endgoodgrief		#skip the "else" part
positive:
	sra	$v0, $a0, 7	#divide by 2⁷ = 128
	mul 	$v0, $v0, $v0	# quotient squared
	and	$v1, $a0, 0x7f  # remainder is rightmost 7 bits
endgoodgrief:
```
There would be other solutions. neg would do here for abs, and of course div would give a quotient and remainder.
The following are pseudo-instructions. Please implement them with assembly language instructions which are not marked with a 'dagger' (and thus correspond to machine language instructions.)

not $v0, $t0
bgt $t0,'@', maybeLetter
```
nor $v0, $t0

slt  $t7, $t0, 'A'	#'A' is character after '@'
beqz $t7, maybeletter   # branch if $t0 is >= 'A' 
      
```

Here is a loop. What do you expect the output to be?

la   $a0, another
li   $v0, 4

loop:	
   syscall		# first time, prints another
   sub  $a0,5		# now a0 points somewhere inside "Help!", probably the e
   sub  $v0,3		# vO is now 1, the print int service
   bne  $v0, 1, loop    # well, it doesn't branch, does it?

li	$v0,10 		# ...and we are finished
	syscall

	.data
string:	  .asciiz	"Help!"
another:  .asciiz  "I need somebody\n"

Well, that wasn't much fun, was it? Had the branch been different, it would have attempted to print an int the second time, and then perhaps an infinite loop?

Suppose that $a0 points to an array of 365 words, containing daily temperatures. Make $v0 be the count of days of extreme temperature, that is, less than -10 or greater than +25

    li $t0, 365	#days of data
    li $v0, 0	#bad day count
tloop:
    beqz $t0, endt	#finished?
	sub  $t0, 1	# one less day to do
	lw   $t1, ($a0)	# get the temperature for this day
	slt  $t2, $t1, -10	#really cold!
	slt  $t3, $t1,  26	#not too hot
	xor  $t2, $t2, $t3	#true if not really cold and not too hot
	bnez $t2, niceday
	    add $v0, 1		# count an extreme day
niceday:
	add  $a0, 4		#point to next array element (don't forget this!)
	j    tloop
endt:				#all done counting.

Write a program in C to add together corresponding elements of 2 arrays of the same size, and print each resulting number, and then the sum of all of them. The data might be defined like this:

#define	SIZE	7
int a[SIZE] = { 1, 3, 5, 7,   9, 88, -1};
int b[SIZE] = {11,-5, 5, 3, -29, -8, 43};

int main() {
	int sum = 0;
	int this;
    for (int i=0; i < SIZE; i++) {  /*go thru arrays*/
	this = a[i] + b[i];	    /*adding the corresponding elements*/
	printf(" %d,",this);
	sum += this;
    }
    printf ("\n It all adds up to %d\n", sum);   /*note newlines */
    return 0;
}
[jensen@shevek temp]$ gcc -std=c99 samp.c 
[jensen@shevek temp]$ ./a.out
 12, -2, 10, 10, -20, 80, 42,
 It all adds up to 132