Would this instruction have worked as well?
or $25,$0,$12 # move s into target register
No — the 32-bit result ($0 OR $12) is moved into $25,
destroying the opcode already there.
$25<——($0 OR $12) # wrong $25<——($25 OR $12) # right
ori $8,$9,0x004A | |||
| 31....26 | 25...21 | 20...16 | 15...................0 |
| opcode | s | d | unsigned const |
| 0xD | 0x9 | 0x8 | 0x004A |
The
same method
of shifting and oring is used with
the bit pattern that designates the destination register.
The const part of the instruction
is in register $13, and it is in the correct field.
It can be copied into $25 using an or instruction.
The complete program follows.
The comments on each line have been improved.
Comments should describe the steps done in solving the problem. They should not just repeat what the instruction does. However, when these notes explain a new instruction the comment may describe the details of the instruction. But that is a poor style, in general.
The following can be copied into the clipboard and pasted into your program editor, then saved to a file and run with SPIM.
## Program to assemble the instruction ori $8,$9,0x004A
##
.text
.globl main
main:
or $25,$0,$0 # clear $25
ori $11,$0,0xD # opcode
ori $12,$0,0x9 # operand $s
ori $13,$0,0x8 # dest. $d
ori $14,$0,0x004A # immediate operand
sll $11,$11,26 # shift opcode into position
or $25,$25,$11 # or it into the instruction
sll $12,$12,21 # shift operand $s into position
or $25,$25,$12 # or it into the instruction
sll $13,$13,16 # shift dest $d into position
or $25,$25,$13 # or it into the instruction
or $25,$25,$14 # or const into the instruction
ori $8,$9,0x004A # The actual assembler
# should create the same machine
# instruction as we now have in $25
## end of file
Would it be possible to dis-assemble the
instruction in $25?