Question

Suppose register $s0 has the binary number of 1111 0000 0000 0000 0000 0000 0000 0000 and the register $s1 has the binary number 00000000 0000 00000000 0000 0000 1111

What the values are of registers $t0 and St1 after these two instructions are executed?
a) SLT $t0, $s0, $s1
b) SLTU St1, $s0, $s1

Answer 1

In assembly language instructions SLT and SLTU of MIPS, $t0 is set to 1 after the 'set on less than' operation, while St1 is set to 0 after the 'set on less than unsigned' operation, due to the binary representations of $s0 and $s1.

Step-by-step explanation:

The question refers to the execution of two assembly language instructions using the MIPS instruction set. The instructions are SLT (set on less than) and SLTU (set on less than unsigned).

For the first instruction SLT $t0, $s0, $s1, it compares the values of registers $s0 and $s1. Since $s0 (1111 0000 0000 0000 0000 0000 0000 0000) is considered to be a negative number in signed comparison (because the highest bit, the sign bit, is 1) and $s1 is positive, $t0 is set to 1, indicating $s0 is 'less than' $s1.

For the second instruction SLTU St1, $s0, $s1, it also compares the values of registers $s0 and $s1, but this time uses an unsigned comparison. In this case, $s0 is larger than $s1, since when comparing binary numbers as unsigned, the value of $s0 is indeed larger.

Therefore, St1 is set to 0, indicating $s0 is not 'less than' $s1 as unsigned integers.