What exactly does the lb instruction do?

Question

I have an exam coming up, and one of the practice problems was:

Assume that $t0 contains the value 0x12121212 and $t1<

Answer 1

The answer would be c) 0xffffff88. The lb instructions sign-extends the byte into a 32-bit value. I.e. the most significant bit (msb) is copied into the upper 24 bits.

0x88 == 0b10001000, i.e. the msb is 1. So the upper 24 bits will be 0b111111111111111111111111 == 0xffffff.