MIPS using different register types but same solution wise

Question

I am wondering is there any difference when you use different register such as the code below:

Here the correct answer uses register $v0 all the way:

get_status:
        lui $t0,0xabab
        ori $t0, $t0, 0x0020

        lw $v0,0($t0)
        andi $v0,$v0,0x4
        srl $v0,$v0,2

        jr $ra

I used $t1 instead of $v0 above, notice I have a extra sw :

get_status:
        lui $t0,0xabab
        ori $t0, $t0, 0x0020

        lw $t1,0($t0)
        andi $t1,$t1,0x4
        srl $t1,$t1,2
        sw $t1, 0($t0)

        jr $ra

I think both code works the same, what you guys think?

Answer 1

$v0 and $t1 are both call-clobbered general-purpose integer registers. They're not "different types".

But in the standard MIPS calling convention, $v0 (and sometimes $v1 ) is where callers expect to find integer return values. That's why get_status calculates a result in $v0 .

I think both code works the same, what you guys think?

Your code modifies the static storage that it loaded from!!! It's not a get_status anymore.

It's either void update_status(void) or int update_status(void) with a non-standard calling convention (returning in $t1 ).

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This is inefficient:

lui ...
ori $t0, $t0, 0x0020
lw  $v0,0($t0)

Use the immediate offset space of lw to hold the low 16 bits of the static address. (It's sign extended, vs. ori zero-extending its immediate, but in this case you don't need to adjust the lui because the offset is signed positive.)

lui ...
lw  $v0, 0x0020($t0)