Three shifts, two questions
You've met all three shift instructions scattered across arithmetic lessons. Time to line them up as what they are — a family answering two questions: which direction, and what fills the vacated bits?
sll — left, zeros fill from the bottom. C's <<.srl — right, zeros fill from the top. C's >> on unsigned values.sra — right, copies of the sign bit fill from the top. C's >> on signed values.
Left shifting has one C spelling, so sll is unambiguous. But >> in C splits into two instructions depending on the operand's type, and that makes right shifts a type oracle, the same way div/divu was. Watch the same shift compiled twice — first on a u32, then on an s32:
srl v0, a0, 6 # u32 x >> 6 — zeros pour in from the top
sra v0, a0, 6 # s32 y >> 6 — the sign bit smears down
Identical C text, different instruction — the declaration alone chose the mnemonic. Decoding runs the oracle backwards: srl in the target forces the shifted value to be unsigned in your C, sra forces it signed. Get it wrong and the diff shows one stubborn mnemonic mismatch on an otherwise perfect line — now you know exactly what that smells like, and the fix is in the declaration, not the expression.
The target below shifts both of its arguments and adds the results. Let each mnemonic pick each parameter's type.
Your task
Write func_8019e9c0, declaring each shifted value with the type its shift demands, to reproduce the target assembly.