Question

.text
main:
   ori $s7,$0,4097       # put a memory location in $s7
   sll $s7,$s7,16       #
   ori $s0,$0,0x2       # put 2 into $s0
and $s0,$s0,$zero # clear $s0
ori $s0,$zero,0x2 # put 2 back into $s0
nor $s0,$s0, $zero # complement $s0
ori $s0,$zero,0x2 # put 2 back into $s0
ori $s1,$0,0x3     # put 3 into $s1
   add $s2,$s0,$s1    # add
beq $s2,$zero, exit
   sw $s2, 4($s7)       # store the result to memory
j exit
ori $s0,$zero,0x2 # pretend like we might do
nor $s0,$s0, $zero # other work after jump

# exit gracefully
exit:   addi $v0,$0,10       # load exit
   syscall           # exit

## end of file

Look at code provided in ALU2015.asm. Write down what it does.

1) What is the result and where is the result stored? Explain how it gets there.

2) Why are they using the ori instruction?

3) Why a complement operation is used in this code?

Answer 1

Assuming it is 32 bit processor.

The main thing that the program is doing is, just moving immediate data into the suitable registers, and then performs addition operation and stores that result in memory.

In this code some dead code is there, those lines of code will never executed. Due to the unconditional jump instruction to exit label.

1)

The final result is 5.

It is stored at location

Because initially $s7 = 4096

i.e., $s7 = 0000 0000 0000 0000 0001 0000 0000 0001

After logical left shifting by 16 bits,

$s7 = 0001 0000 0000 0001 0000 0000 0000 0000

The result is storing at 4($s7) i.e., 4 + $s7

Result location = 0001 0000 0000 0001 0000 0000 0000 0100

2)

ori instruction is used for moving the immediate data into the registers. It will take less time to move the data when compared to mov instructions.

3)

Here the complement operation is used to complement the data in $s0 to one's complement form. But after that that register is modified.

And another complement is used in dead code, which cannot executed by the assembler.