Hallo,
MIPS works with words of 32 bit. But for universitiy, i have to add two 64 bit Integers numbers.
first of all, my idea was that i split the numbers and store them in two register. But that's only a theoretical idea. I have no idea how to realize in MIPS Code. How can i split the number? Anybody an idea?
Thank you very much
Hello,
greetings from the closest hood !
Which mips assembly?
The idea you sketched out is very good! It only functions this way.
# Assuming that the 64 bits int are always in two 32 bit regs ($11, $10) and ($21, $20),
# where $11 and $21 contain the high order parts, the addition can be made by
# 1) unsigned add both low order parts $10 + $20, result in $5
addu $5, $10, $20
# Then you need an add with carry instruction,
# unfortunately mips assemblies lack of such nice instruction.
# 2) So you must make use of sltu instruction, set carry in $4:
sltu $4, $5, $20 # fast way to determine carry of $5 = $10 + $20
# 3) Now add with carry can be carried out:
addu $4, $4, $11 # carry + first high order part $11
addu $6, $4, $21 # intermediate result + second high order part $21, result in $6
# 4) Finally, did we even do a 65 bit addition?
sltu $4, $6, $21 # carry of last addu set in $4.
### Final result is:
### ($4, $6, $5), where $4 is last carry, $6 high order and $5 low order part of 64 bit result.
# Attention: Above assembly not tested! If you find any error, pls tell me.Well, that's all what I could remember of vintage mips r3000, hopefully, that some more advanced R.I.P. mips machines also function similar to this void-of-add-with-carry era, or they had even got that missing instruction, what you should really check for your current mips machine.
Happy mipsing!
-- tesu
so, so, doing excercises at LMU
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