Question

How do I solve this problem? Please, show steps.

Problem: Register s1 contains the value 23, and register s2 contains the value 30. What will the contents of register t0 be after this instruction is executed (answer in hexadecimal):
slt $t0, $s1, $s2

Answer 1

Slt:

• It is an instruction used in MIPS programming for comparing the contents between two registers. The full form of slt is Set on Less than.
• It sets the resultant register to 1 if the first register operand is less than the second register operand otherwise it is 0 which is shown in the below example:

slt $t0, $t1, $t2   # The register $t0 is 1 when the register value $t1<$t2 Otherwise 0

Where $t0 is the resultant register

$t1 is the first register

$t2 is the second register

Consider the register values as given in the question.

$s1, 23h                         # Register S1 contains the value 23.

$s2, 30h           # Register S2 contains the value 30.

slt $t0, $s1, $s2   # $t0=1 because $S1 is less than $S2 (23<30).

Therefore, after executing the instruction, the contents of the register t0 is 1 (0001h).