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Understanding MIPS
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I have a block of MIPS code that implements a pseudo-instruction operating on the values from two registers. The problem is i am having trouble undertanding what it does. The code is:
sll $t0, $s0, 31
srl $s0, $s0, 1
srl $s1, $s1, 1
or $s1, $s1, $t0
I think it is a divide by 2 checker? If any one can help me out with what it does, that would be great.
Cheers
sll $t0, $s0, 31
srl $s0, $s0, 1
srl $s1, $s1, 1
or $s1, $s1, $t0
I think it is a divide by 2 checker? If any one can help me out with what it does, that would be great.
Cheers
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Originally Posted by NZStudent 
I have a block of MIPS code that implements a pseudo-instruction operating on the values from two registers. The problem is i am having trouble undertanding what it does. The code is:
I think it is a divide by 2 checker? If any one can help me out with what it does, that would be great.sll $t0, $s0, 31 srl $s0, $s0, 1 srl $s1, $s1, 1 or $s1, $s1, $t0
Cheers
sll $t0, $s0, 31
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....
srl $s0, $s0, 1
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...
srl $s1, $s1, 1
whatever is in $s1 will be shifted right 1 placed back into $s1
or $s1, $s1, $t0
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
Here is his response
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Mips assembler is in the formsll $t0, $s0, 31 srl $s0, $s0, 1 srl $s1, $s1, 1 or $s1, $s1, $t0
opcode destination source2 source1
Haven't used the MIPS but I do have the manual.
It's a macro
$t0 is a temp register
$s0 and $s1 will be source registers for a 64 bit value. $s0 is the
high 32 bits.
Puts the low bit into the top bit position of $t1 assuming the shiftsll $t0, $s0, 31
count is one based.
s0 is shift right one and top bit position is zero.srl $s0, $s0, 1
Remove the bottom bit of $s1srl $s1, $s1, 1
or in bottom bit from $s0or $s1, $s1, $t0
It's a 64bit right shift $s1 is the low 32 bits, $s0 is the high 32 bits.
Last edited by Paul.Esson; Oct 7th, 2006 at 9:13 am.
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
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
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