li $v0,12 == ori $v0,$zero,12
The load immediate pseudoinstruction is useful for loading a register with an integer value. Here is another example:
li $t2,-156
This puts the two's complement representation of -15610
into register $t2.
However, this can not be done with
ori
because the immediate field of that instruction must be a
16-bit unsigned integer
(which is zero-extended to 32 bits upon execution).
Because the immediate value of the
li
is negative, a different translation is used:
li $v2,-156 == addiu $v2,$0,-156
(Remember that addiu
sign-extends its immediate operand before
performing the addition.
Also remember that the u part
of the mnemonic means that overflow will not cause
a trap.
The addiu instruction can be used
with negative integers).
The extended assembler translates
li
into different basic instructions depending on the sign of the immediate value.
Can either basic instruction,
ori
or
addiu,
use an immediate value greater than 16 bits?