lui $13, 0x0060 ori $13, $13, 0x5000 lw $12, 0xC($13)
The
above ori instruction "fills in" the
lower 16 bits of register $13 by doing the following:
$13 after lui : 0000 0000 0110 0000 0000 0000 0000 0000 zero-extended imm. op. : 0000 0000 0000 0000 0101 0000 0000 0000 result of bitwise OR : 0000 0000 0110 0000 0101 0000 0000 0000
Other sequences of instructions also will work. Choose whichever method works best in the context of the rest of the program.
Because the "upper half" of an address is 16 bits and
the offset of the lw can hold the 16 bit "lower half,"
the two in combination can
address any byte of memory.
The problem was to load $12 with the word at address
0x0060500C.
Here is another way to do it:
Split the address into halves: 0x0060 and 0x500C.
Load the top half into $13 and use the bottom half
as the offset.
lui $13, 0x0060 lw $12, 0x500C($13)
Since the ori instruction is used often with
the destination register as one of the operands,
there is a
shorthand instruction
in assembly language.
ori $d,const # assembles to
# the same machine
# instruction as
# ori $d,$d,const
Do the following two assembly instructions assemble to the same machine instruction?
ori $10,$10,0x00C4 ori $10,$0, 0x00C4