I have little knolage of assembler but here is my thoughts

We are shifting whatever is in $s0 left 31 places, assuming we have a 32 bit register.. this will mean that if the lowest bit is 1 then the result will be
10000000000000000000000000000000 and if the lowest bit is 0 then result 00000000000000000000000000000000 this will be put in the register dictated by $t0....


shift whatever is in $s0 right 1 and assign it to $s0.. I don't see $s0 being used again.. therefor don't have a clue why we would do this...


whatever is in $s1 will be shifted right 1 placed back into $s1


or the two values together then place them back into $s1.
so since we have shifted s1 right one, Its highest bit will be 0, therefor the highest bit of the future s1 depends on the lowest bit of $s0.

Thefore what this appears to do, is if s0 is even, the resulting number will will have 0 in the highest bit, while if s0 is odd the resulting number will have 1 in the highest bit.....

why we would do this, I don't know... But please somone enlighten me

After I was completly wrong, I sent an email off to my dad, He has some experiance programming in assembler
Here is his response

sll $t0, $s0, 31
srl $s0, $s0, 1
srl $s1, $s1, 1
or $s1, $s1, $t0

It's a macro fragment and the macro header would help.

$t0 is a temp register, $S0 and $S1 are a 64 bit value ( $S0 is the high 32 bits, $S1 low 32 bits. After excecution the 64 bit value will be logicaly shifted right one bit possition.

It works lik this.
sll $t0, $s0, 31 Put bit 0 of the high bits into bit position 31 in temp.
srl $s0, $s0, 1 High bits right 1 bit position
srl $s1, $s1, 1 Low bits right 1 bit position
or $s1, $s1, $t0 Or in the bit shift out of the high register

Regards

Opps I didn't see that you had posted the answer.