Shifting the Opcode

The ori instruction puts its immediate operand in the low order bits (the right-most bits) of its destination register. So now, with the following code, the registers have the correct patterns but the patterns are not in the correct fields.

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

Register $11 contains the opcode in its right-most bits, like this:

31....26	25...21	20...16	15...................0
000000	00000	00000	0000 0000 0000 1101

However, the opcode is required to be in the high-order six bits, 26 through 31. A sll will shift the contents of $11 into the correct position.