QUESTION 12:
How could you compute 4 × $8 + $8 ?
addiu $8,$0,-86
Yes. The immediate operand -86 is sign-extended to 32 bits then added to a 32-bit zero. The sum (-86) is loaded into $8.
You would expect that since there are
add, addu, addi, addiu
and since there are sub, subu
that there would be subtract immediate instructions.
But there are not.
The add immediate instruction is used.
To subtract 201
from register $10 using an immediate operand,
do this:
addiu $8,$10,-201 # $8 <-- $10 - 201
Say that we want to compute 5 × x - 74 where the
value x is in register $8.
MIPS has an integer multiply instruction,
but let us say that we don't want to use it.
How can 5 × x be done using the instructions you have seen
so far?
How could you compute 4 × $8 + $8 ?