Assignment 2 MIPS instructions


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Assignment 2

Note: Make reasonable assumptions where necessary and clearly state them. Feel free to
discuss problems with classmates, but the only written material that you may consult while
writing your solutions are the textbook and lecture slides/videos. Solutions should be
uploaded as a single pdf file on Canvas. Show your solution steps so you receive partial
credit for incorrect answers and we know you have understood the material. Don’t just
show us the final answer.
1. Annotate the following MIPS instructions to indicate source registers and destination
registers. A source register is read during the instruction’s execution, while a
destination register is written during the instruction’s execution. (12 points)
1. add $s1, $t2, $t3
2. lw $s3, 8($gp)
3. sw $s4, 12($s5)
4. addi $s1, $zero, 100
5. bne $t1, $t2, else
6. add $s1, $s1, $s1
2. Consider a program that declares global integer variables x, y[10]. These variables are
allocated starting at a base address of decimal 1000. All these variables have been
initialized to zero. The base address 1000 has been placed in $gp. The program
executes the following assembly instructions:
lw $s1, 0($gp)
addi $s1, $s1, 25
sw $s1, 0($gp)
lw $s2, 12($gp)
add $s2, $s2, $s1
sw $s2, 8($gp)
sw $s2, 12($gp)
1. What are the memory addresses of variables x, y[0], and y[1]? (15 points)
2. What are the values of variables x, y[0], y[1], and y[2] at the end of the program?
(20 points)
3. Express the following decimal number in binary and hexadecimal forms: 146. (6
4. Express the following binary number in decimal and hexadecimal forms: 1001100. (6
5. Express the following hexadecimal number in decimal and binary forms: 0x6d. (6
6. Write the MIPS assembly code that corresponds to the pseudo code below. Assume
that the address for integer i is baseaddress+4 and the address for a[0] is
baseaddress+8. Assume that the baseaddress is stored in $gp. The code initializes i to
0; it then iterates from i=0 to i=9, setting a[i] = 4i in each iteration. To make your
code efficient, i must be loaded into a register at the start, and it must be updated in
memory only after you’ve finished the for loop.
for (i=0; i<10; i++)
2018/10/7 CS/EE 3810 2/2
a[i] = 4*i;
(35 points)

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