sll $12,$12,21 # left shift s by 21 places or $25,$25,$12 # move s into target register
The
low-order bit of the pattern for $s is in bit position
0 (of register $12). It must be moved to position 21 of
the instruction.
This is done by shifting left by 21 places.
Next, the or copies it into
$25, where the instruction
being assembled.
The or works like this:
| 31....26 | 25...21 | 20...16 | 15.............................0 | |
| Register $12 (before shift) | 000000 | 00000 | 00000 | 0000 0000 0000 1001 |
| Register $12 (shifted) | 000000 | 01001 | 00000 | 0000 0000 0000 0000 |
| Register $25 (before or) | 001101 | 00000 | 00000 | 0000 0000 0000 0000 |
| Register $25 (after or) | 001101 | 01001 | 00000 | 0000 0000 0000 0000 |
Would this instruction have worked as well?
or $25,$0,$12 # move s into target register