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After the instruction ori $8,$0,0x2 executes, what is in $8?

Answer:

The 16-bit 0x2 immediate operand has been zero-extended and copied into register $8.


OR Immediate Instruction

ori  d,$0,const    # register d  ← const.
                   # const is 16-bits, so
                   # 0x0000 ... const ... 0xFFFF

If the OR operation is done with the zeros in register $0, the result is a copy of the zero-extended immediate operand into the result register.

Copying a bit pattern into a register is usually called loading the register. Register $8 was loaded with a 32-bit pattern. The pattern could represent a positive two. If so, register $8 was loaded with positive two.

If const represents an integer, then 0 ≤ const ≤ 65535.

Of course, the operation is pure bit manipulation and the bits might represent something other than an integer.

The three operands of the assembly instruction d, $0, and const must appear in that order.


QUESTION 4:

Can the immediate operand of an ori be regarded as a signed integer?


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