Question

assuming that the memory location 0x10010000 contains the value 0x00000080 and the mips-32 processor uses the little-endian scheme. after a mips-32 processor executed the following instructions: li $s0, 0x10010000 lb $t0, ($s0) the content of register $t0 is 0x[a]

Answer 1

After executing the instructions "li $s0, 0x10010000" and "lb $t0, ($s0)" on a MIPS-32 processor with the little-endian scheme, the content of register $t0 would be 0x80.