CS 271 Homework 2
Fall 2008
IPFW
Exercises 2.2, 2.6, 2.20, 2.30, 2.38
2.2 By lookup using the table in Figure 2.5 on page 62,
7fff fffahex = 0111 1111 1111 1111 1111 1111 1111 1010two
= 2,147,483,642ten
2.6
sll $t0, $t3, 9 # shift $t3 left by 9, store in $t0
srl $t0, $t0, 15 # shift $t0 right by 15
2.20 Figure 2.21 shows decimal values corresponding to ACSII characters.
2.30
sll $a2, $a2, 2 # max i= 2500 * 4
sll $a3, $a3, 2 # max j= 2500 * 4
add $v0, $zero, $zero # $v0 = 0
add $t0, $zero, $zero # i = 0
outer: add $t4, $a0, $t0 # $t4 = address of array 1[i]
lw $t4, 0($t4) # $t4 = array 1[i]
add $t1, $zero, $zero # j = 0
inner: add $t3, $a1, $t1 # $t3 = address of array 2[j]
lw $t3, 0($t3) # $t3 = array 2[j]
bne $t3, $t4, skip # if (array 1[i] != array 2[j]) skip $v0++
addi $v0, $v0, 1 # $v0++
skip: addi $t1, $t1, 4 # j++
bne $t1, $a3, inner # loop if j != 2500 * 4
addi $t0, $t0, 4 # i++
bne $t0, $a2, outer # loop if i != 2500 * 4
The code determines the number of matching elements between the two arrays and returns this
number in register $v0.
2.38 The problem is that we are using PC-relative addressing, so if that address is
too far away, we won’t be able to use 16 bits to describe where it is relative to the
PC. One simple solution would be
here: bne $s0, $s2, skip
j there
skip:
…
there: add $s0, $s0, $s0
This will work as long as our program does not cross the 256MB address boundary described in the
elaboration on page 98.