What fills the top: nothing
The unsigned byte answers the 24-bit question differently: fill with zeros. That's lbu — load byte unsigned — and this time the u means exactly what it says. Here's fetch, returning the third byte of a u8 array:
s32 fetch(u8 *src) {
return src[2];
}
lbu v0, 2(a0) # v0 = src[2], zero-extended: top 24 bits are 0
jr ra
nop
The same memory byte 0xFB now arrives as 0x000000FB — 251, not -5. Same bits in memory, two different 32-bit numbers in a register; the declaration decides which, and the mnemonic broadcasts the decision:
lb — the top bits mirror bit 7. Value range -128…127. C type s8.lbu — the top bits are zero. Value range 0…255. C type u8.
When you're staring at a target, run the oracle backwards: every lbu forces a u8 somewhere in your declarations, every lb an s8. Get one wrong and the diff disagrees on exactly that mnemonic — the register choreography around it will match perfectly, which is precisely how you'll know the type is the bug, not the expression.
The target reads one byte through each of its pointers and combines them — a one-line function, but only if each parameter is declared the way its load mnemonic demands.
Your task
Write func_802319c0 to reproduce the target assembly.