Description
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
points)
4. Express the following binary number in decimal and hexadecimal forms: 1001100. (6
points)
5. Express the following hexadecimal number in decimal and binary forms: 0x6d. (6
points)
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++)
CS/EE 3810
http://www.cs.utah.edu/~rajeev/cs3810/hw/hw18-2.html 2/2
a[i] = 4*i;
(35 points)
