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

### A good answer might be:

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

# OR Immediate Instruction

Because the OR operation was done with the zeros in register $0, the result was a copy of the zero-extended immediate operand. 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. Here is a description of the `ori` instruction when used to load a register:

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

If `const` represents an integer, then 0 <= `const` <= 65535. 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?