ori $8, $0, 0x6F # put bit pattern register into $8
sll $8, $8, 2 # shift left logical by two
Yes, this is OK.
When an ALU operation is performed:
There is no problem when an operand register is also the result register because the operand data was transferred to the ALU in the first step, leaving the register open to receive the result.
The assembly language statement in the example program is:
sll $9, $8, 2 # shift left logical by two
Here is the machine instruction corresponding to that statement:
0 0 0 8 4 8 8 0 -- machine instruction in hex 0000 0000 0000 1000 0100 1000 1000 0000 -- machine instruction in bits 000000 00000 01000 01001 00010 000000 -- fields of the instruction opcode ----- source dest shft 2ndary -- meaning of the fields ALUop $8 $9 2 sll -- fields decoded
The sll operation is designated
by the 6-bit zero at the beginning
of the instruction and the 6-bit zero at the end.
The remaining fields give the source register ($8 in this case),
the destination
register ($9 in this case),
and the number of positions to shift (2 in this case).
Notice that in the machine
language the register designators are reversed
from how they occur
in the assembly language.
The five zero bits after the opcode are not used. All MIPS instructions are 32 bits. These bits are arbitrarily given the value zero.
Study the above machine instruction.
Write the machine instruction for
shift-left-logical the pattern in $0
by zero positions and leave the result in $0.
Here is the instruction in assembly language:
sll $0, $0, 0 # fun machine instruction
Write your machine instruction (as a bit pattern) here:
Hint: on the midterm, I expect you to write the 32-bit pattern for this instruction without hesitation! Memorize it now!