Download Homework 1 | Computer Systems and Assembly Language | EECS 388 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
EECS 388: Computer Systems and Assembly Language
Homework 1
Due Feb. 12, 2007 Justify your answers!
Figure 1 shows a part of the memory (both contents and locations). Contents Locations
$20 $ $50 $ $01 $
$B5 $
$CD $
Figure 1.
Problem 1 (10 Points): Consider the memory shown in Figure 1. Write a program to add the two numbers in $4000 and $4001 and store the results in $5002.
Here, we simply need to add two 8-bit numbers and store the 8-bit result, so we can use accumulator A for this task:
ORG $ LDAA $ ADDA $ STAA $ SWI
Problem 2 (10 Points): Consider the memory shown in Figure 1. What is the content in accumulator D after the following lines are executed?
LDD $ ADDA $
First we load D with the 16-bit value starting at location $4001. At this point, we have
D: $
Next we add the 8-bit value at location $5000 to the value contained in A (which is the most significant byte of D, $50):
A: $50 + $B5 = $05 (The real result is $105)
The above result is too large to fit in A, so A will contain $05 and the overflow bit in the CCR will be set. Consequently, the upper byte of D (which is A) is now $05 but the lower byte has not changed at all, giving us:
D: $
Problem 3 (30 points): Consider the memory shown in Figure 1. What are the contents in memory location $5002 and accumulator A after the following lines are executed?
LDAA $ LDD #$ LDAB $ STD $
First, we load accumulator A with the 8-bit value at location $4000, which gives us:
A: $
Next, we load D with the immediate 16-bit value $5000, which now gives us:
D: $
But since D is really A and B, this means:
A: $ B: $
Next we load accumulator B with the 8-bit value at location $4002, which gives us:
B: $
So now we have:
A: $ B: $ D: $
Finally we store the 16-bit value in D to memory location, starting at location $5002, so now memory will look like:
Contents Locations
$20 $ $50 $
c) LSLD
Original(hex): $ B B C D Original(bin): 1011 1011 1100 1101 Shifted(bin): 0111 0111 1001 1010 Shifted(hex): $ 7 7 9 A
Final answer:
A: $77 B: $9A
d) ROLB
Remember that B is the lower byte of D, and that the carry bit is 1 before the rotate:
Original(hex): $ B B C D Original(bin): 1011 1011 1100 1101 Shifted(bin): 1011 1011 1001 1011 Shifted(hex): $ B B 9 B
The upper byte of D (which is A) is not affected by the rotate. Final answer:
D: $BB9B
Problem 4 (10 Points): Write a program to add two values $20 and $40 and store the result in memory location $8000.
We are asked to add to immediate 8-bit values and store the 8-bit result in memory.
ORG $ LDAA #$ ADDA #$ STAA $ SWI
Problem 5 (15 Points): Write a program to implement the following 16 bit rotation, i.e., originally, A and B contain a 7 a 6 …a 0 and b 7 b 6 …b 0 , respectively. After the 16 bit rotation, the contents in A and B are a 6 a 5 …a 0 b 7 and b 6 b 5 …b 0 a 7 , respectively. (Note: You can only use the data transfer and manipulation instructions in Section 2.4.1).
There are a lot of ways to do this problem; here are a few solutions (we only show the instructions for brevity):
Solution 1:
ASRA ROLA ROLB ROLA
Solution 2:
ASRB ROLB ROLA ROLB
Solution 3:
STAB $ LSLD LDAB $ ROLB
Solution 4:
STAA $ ROLA ROLB LDAA $ ROLA
Solution 5:
