converting C code to MIPS Assembly Language with arrays

Question

Ok, so I have to convert the following C code segment to MIPS Assembly.

f = k + A[5]

The question tells me that f is stored in register $s3, k is in $s2 and the base address of array A is $s4. This is what I put as my answer:

add $s3, $s2, $s4

Is this correct? Do I have to do anything special with the 5 in the array? I'm very new to MIPS, so any and all help if VERY much appreciated.

"converting C code to MIPS" - An entire family of computer programs is designated to that task. — user529758
– user529758, Commented Apr 8, 2013 at 19:47

Mercurybullet · Accepted Answer · 2013-04-09 22:28:24Z

1

Are you working on this for homework? If so, are you actually writing out an executable program or just responding to a list of questions?

Either way yes, you do need to account for the 5 in the array. The question is telling you that $s4 points to the base address of the array, not the 5th index.

hint: A[0] would be at the same address as the base of the array.

edited Apr 9, 2013 at 22:28

answered Apr 8, 2013 at 19:46

Mercurybullet

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2 Comments

Mercurybullet Over a year ago

@wazy thanks for the heads up. I noticed nothing came up when I went searching to retag, but didn't find anything. Is there a new tag for homework type questions or does SO condone just giving away hw answers these days?

Mercurybullet Over a year ago

For anyone interested, I found this link on meta. Indeed the homework tag has been deprecated. Editing my response to more politely ask if the question is homework, but still only guiding towards correct answer.

wazy · Accepted Answer · 2013-04-08 20:08:54Z

0

Try this out. (Off the top of my head). Remember each index is * 4.

li $t2, 6            # init 6 to $t2
addi $t2, $t2, $t2   # $t2 * 2
addi $t2, $t2, $t2   # $t2 * 2
addi $t1, $t2, $s4   # A[6 * 4] 
lw $t4, 0($t1)       # load A[6] int $t4
addi $s3, $s2, $t4   # obtain f

edited Apr 8, 2013 at 20:08

answered Apr 8, 2013 at 19:52

wazy

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