Answer:
No. Because the 16-bit pattern is zero-extended to 32 bits, the resulting pattern is never a two's-complement negative integer.
No. Because the 16-bit pattern is zero-extended to 32 bits, the resulting pattern is never a two's-complement negative integer.
Here
is a description of the ori instruction used
as a bitwise OR instruction:
ori d,s,const # register d <-- bitwise OR ofconst# with the contents of register $s #constis 16-bits, so # 0x0000 ...const... 0xFFFF
The parts of the instruction must appear in the correct order,
and const must be within the specified range.
If the immediate operand in
the assembly language shows less than 16 bits
(as does 0x2 in the previous example)
then the assembler expands it to the required sixteen bits.
If the assembly language
specifies more than sixteen bits,
then the assembler writes an error message.
The const part of the assembly language instruction
can be a positive decimal or a hexadecimal constant.
The assembler translates the constant into a 16-bit pattern in
the machine instruction.
For example, the following two assembly language instructions
translate into the same machine language instruction:
ori $5,$4,0x10 ori $5,$4,16
Is the following instruction correct?
ori $0,$9,0x32