Register s1 contains the value 32, 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
Register s1 contains the esteem 32, and register s2 contains the esteem 30. What gain the solution of register t0 be succeeding this information is performed (exculpation in hexadecimal):
slt $t0, $s1, $s2
Exculpation is as follows:
$ s1 = 32
$ s2 = 30
slt , fixed on near than (signed) information states that
now according to slt $t0, $s1, $s2 information of MIPS information fixed if $s1 is near than $s2, than $t0 is fixed to 1 else if it is senior than esteem of $t0 is fixed to 0.
Now
slt $t0, $s1, $s2
= slt $t0, 32,30
so the esteem of $t0 in the information is fixed to binary ‘0’
and HEXADECIMAL equivalent of BINARY ‘0’ is ‘0′