Understanding Immediates in ADDI and similar NIOS II instructions

As we discussed in the lecture, ADDI takes two source operands, one being a register and the second being a 16-bit immediate (number). That immediate get’s sign-extended into 32-bits before participating in the addition.

Let’s look at a couple of examples:

What is the result of:

1. addi r10, r0, 0x8000?

a. R10 = 0xFFFF8000

2. Addi r10, r0, 0xAAAA?

a. R10 = 0xFFFFAAAA

3. Addi r10, r0, 0x9000?

a. R10 = 0xFFFF9000

4. Addi r10, r0, 0x7AAA?

a. R10 = 0x00007AAA?

5. Addi r10, r0, -2?

a. -2 = 0xFFFE in hexadecimal and in 16 bits

b. R10 = 0xFFFFFFFE, this is -2 in hexadecimal in 32 bits

c. Note: 0x0000FFFE is not -2, it’s 65534

6. Addi, r10, r0, 32769?

a. Got to convert 32769 in binary. Turns out it’s 1000 0000 0000 0001, or 0x8001 in hexadecimal

b. So, r10 = 0xFFFF8001

7. What will be r10 after these two instructions execute?

a. Addi r10, r0, 0xFFFF

b. Addi, r10, r10, 1

i. You should get r10 = 0

8. How about these two?

a. Addi r10, r0, 3

b. Addi r10, r10, 0xFFFE

i. You should get r10=1

9. These two?

a. Movhi r10, 0xFFFF

b. Addi r10, r10, 0xFFFF

i. You should get r10 = 0xFFFEFFFF

ii. Try everything in binary and see what you get

10. These?

a. Movhi r10, 0x1234

b. Addi r10, r10, 0x5678

i. You should get r10=0x12345678

11. Finally, these two:

a. Movhi r10, 0x1234

b. Addi r10, r10, 0x89ab

i. You should get r10=0x123389AB

Please double-check these and let me know of any errors.