Exam 2 Review Questions - Part 2 (with Answers)

This is representative of the kinds of topics and kind of questions you may be asked on the exam. In addition to this practice exam, you should also review:

The handouts
Labs

Questions

Convert the following C-like code into MIPS assembly. The names of the variables reflect which registers must be used for the MIPS assembly. Do not assume any initial values for the registers. You may use additional registers.
```
if ($t0 == 0) {
	$t1 = 5;
}
```
Answer:
```
  bne $t0, $zero, after_if
  li $t1, 5
after_if:
```
Convert the following C-like code into MIPS assembly. The names of the variables reflect which registers must be used for the MIPS assembly. Do not assume any initial values for the registers. You may use additional registers.
```
if ($t0 < 5) {
	$t1 = 0;
} else {
	$t1 = 1;
}
```
Answer:
```
  slti $t3, $t0, 5
  beq $t3, $zero, else_branch
  li $t1, 0
  j after_if
else_branch:
  li $t1, 1
after_if:
       
```
Convert the following C-like code into MIPS assembly. The names of the variables reflect which registers must be used for the MIPS assembly. Do not assume any initial values for the registers. You may use additional registers.
```
if ($t0 == 0 || $t1 == 1) {
	$t2 = 5;
} else {
	$t2 = 6;
}
```
Answer:
```
  beq $t0, $zero, true_branch
  li $t3, 1
  beq $t1, $t3, true_branch
  li $t2, 6
  j after_if
true_branch:
  li $t2, 5
after_if:
       
```

Convert the following Java/C-like code into MIPS assembly. The names of the variables reflect which registers must be used for the MIPS assembly.

if ($t0 >= 0 && $t0 < $t1) {
	$t2 = 9;
} else {
	$t2 = 0;
}

Answer:

# !($t0 >= 0 && $t0 < $t1) ==> $t0 < 0 || $t0 >= $t1
  slt $t3, $t0, $zero # $t0 < 0?
  bne $t3, $zero, else_branch # if yes, jump to else
  # $t0 >= $t1 ==> !($t0 < $t1)
  slt $t3, $t0, $t1 # $t0 < $t1?
  beq $t3, $zero, else_branch # if not, jump to else
  li $t2, 9
  j after_if
else_branch:
  li $t2, 0
after_if:

Write a MIPS program that will read integers from the user until 0 is input. Once 0 is input, the program should print the sum of all the numbers read in. As a hint, you should track a running sum, instead of trying to store all the numbers the user read in. If the user immediately inputs a 0, then the running sum should be 0.

Answer:

.text
main:
	# $t0: running sum
	# $t1: input number
	li $t0, 0
	# $t1 = readNum()
	# while ($t1 != 0) {
	# $t0 = $t0 + $t1
	# $t1 = readNum()
	# }
	# print($t0)
	
	li $v0, 5
	syscall
	move $t1, $v0
loop_begin:
	beq $t1, $zero, loop_end
	add $t0, $t0, $t1
	li $v0, 5
	syscall
	move $t1, $v0
	j loop_begin
loop_end:
	move $a0, $t0
	li $v0, 1
	syscall
	
	li $v0, 10
	syscall

$t0 = 10
$t1 = 1
$t2 = 0
while ($t1 <= $t0) {
$t2 = $t2 + $t1;
$t1++;
}

Answer:

.text
main:
	li $t0, 10
	li $t1, 1
	li $t2, 0
	loop_begin:
	# $t1 <= $t0 ==> !($t1 > $t0) ==> !($t0 < $t1)
	slt $t3, $t0, $t1
	# $t3 == 0 if $t0 < $t1, meaning !($t1 > $t0), meaning $t1 <= $t0
	# $t3 == 1 if !($t1<= $t0)
	bne $t3, $zero, loop_end
	add $t2, $t2, $t1
	addi $t1, $t1, 1
	j loop_begin
loop_end:
	li $v0, 10
	syscall

Write a MIPS program that will read in an integer, and will print one of two things:

Bit 2 is set
Bit 2 is not set

...depending on whether or not bit 2 of the input number is set. To be clear, bit 0 refers to the rightmost bit in the number.

Answer:

.data
is_set_string:
	.asciiz "Bit 2 is set\n"
is_not_set_string:
	.asciiz "Bit 2 is not set\n"
.text
main:
	# $t0: the number
	# read in the integer
	li $v0, 5
	syscall
	move $t0, $v0
	# mask out all other bits
	andi $t0, $t0, 0x4
	beq $t0, $zero, is_not_set
	la $a0, is_set_string
	j after_if
is_not_set:
	la $a0, is_not_set_string
after_if:
	# print the string
	li $v0, 4
	syscall
	# exit the program
	li $v0, 10
	syscall

int s0 = 82;
int s1 = s0 >> 2;
int s2 = s1 * 20;
int s3 = s2 + 7;
int s4 = s3 - 24;
int s5 = s4 / 3;

Answer:

main:
	li $s0, 82 # int s0 = 82;
	
	sll $s1, $s0, 2 # int s1 = s0 << 2;
	
	li $t0, 20 # int s2 = s1 * 20 (part 1 of 3)
	mult $s1, $t0 # (part 2 of 3)
	mflo $s2 # (part 3 of 3)
	
	addi $s3, $s2, 7 # int s3 = s2 + 7
	
	li $t1, 24 # int s4 = s3 - 24 (part 1 of 2)
	sub $s4, $s3, $t1 # (part 2 of 2)
	
	li $t2, 3 # int s5 = s4 / 3 (part 1 of 3)
	div $s4, $t2 # (part 2 of 3)
	mflo $s5 # (part 3 of 3

Convert the following C-like code into MIPS assembly. The names of the variables reflect which registers must be used for the MIPS assembly. Do not assume any initial values for the registers. You may use additional registers. The portions in <<>> will require you to use QtSpim functionality. You do not need to exit the program properly.

int s0 = <<read integer from the user>>;
int s1 = 2;
if (s0 < 7) {
s1 = 3;
}
<<print integer s1>>
if (r2 < r3 && r3 < r4) {
  r5 = r6;
} else {
  r6 = r5;
}

Answer:

main:
	# read integer from user
	li $v0, 5
	syscall
	# save integer from user
	move $s0, $v0
	# store 2 in s1
	li $s1, 2
	# check if s0 < 7
	li $t0, 7
	slt $t1, $s0, $t0
	# if it's NOT less than 7, skip the body of the if
	beq $t1, $zero, printmsg
	# we didn't branch, meaning s0 < 7
	li $s1, 3
printmsg:
	# print s1
	li $v0, 1
	move $a0, $s1
	syscall

int s0 = <<read integer from the user>>;
int s1 = 2;
if (s0 < 7) {
s1 = 3;
} else {
s1 = s0 + s0;
}
<<print integer s1>>

Answer:

main:
	# read in the integer from the user, and initialize s1
	li $v0, 5
	syscall
	move $s0, $v0
	li $s1, 2
	# check if $s0 < 7
	li $t0, 7
	slt $t1, $s0, $t0
	# jump to the else branch if this isn't true
	beq $t1, $zero, else_branch
	# fall through to the true branch
	li $s1, 3
	j print
else_branch:
	add $s1, $s0, $s0
	# fall through to the print
print:
	li $v0, 1
	move $a0, $s0
	syscall

int s0;
int s1 = 1;
for (s0 = 0; s0 < 10; s0++) {
	s1 = s1 * s0;
}

Answer:

main:
	# initialize variables
	li $s0, 0
	li $s1, 2
loop:
	# check loop condition
	li $t0, 10
	slt $t1, $s0, $t0 # s0 < 10?
	beq $t1, $zero, loop_exit # if not, jump to loop_exit
	# do body of the loop
	mult $s1, $s0
	mflo $s1
	# increment counter
	addi $s0, $s0, 1
	j loop
loop_exit:
	# this is past the loop