Bit 31 is the sign
For a signed 32-bit value, the top bit is the sign — 1 means negative. Shift right by 31 and that single bit is all that's left, which makes "is this negative?" a one-instruction question. But which right shift you use changes the answer's shape. Signed first:
sra v0, a0, 31 # 0 if x >= 0, else 0xffffffff
jr ra
nop
sra drags copies of the sign bit down the whole register: the result is 0 or all-ones — a mask, ready to AND with something. That's x >> 31 on a signed x. Now the unsigned shift:
srl v0, a0, 31 # 0 if x >= 0, else 1
jr ra
nop
srl pours in zeros instead, leaving just 0 or 1 — a boolean: "the sign bit, as a flag". Since srl is the unsigned shift, the C has to shift an unsigned value — a cast, (u32)x >> 31, on an x that's otherwise signed.
Both shapes are all over compiled code, and the mnemonic tells you instantly which one you're looking at: sra 31 builds a sign mask, srl 31 builds a sign flag.
The target below computes something from its arguments first, then takes the sign flag of the result — a two-step read: decode the arithmetic, then hang the cast-and-shift off it.
Your task
Write func_801bd184 to reproduce the target assembly.