Quiz submission record for quiz2-1-2 at Tue Jun 29 00:43:51 2004:

Your Answer for Question 1:
In MIPS there are only 32 registeres, so since programs have many variables, it cannot be directly translated in assembly code. Also since MIPS words start at multiples of 4, A[8] in MIPS addressing scheme would be A[8/4], so you would want to multiply the C address by 4 to get the correct address in MIPS, which is 4x8=32.


Your Answer for Question 2:
add $k,$k,1

Your Answer for Question 3:
i did not know that MIPS used a method similar to slab allocation, in which addresses are split up into words starting at multiples of 4.


Your unique submission ID is quiz2-1-2-cs61c-av-1088495031-476.