Convert MIPS32 to C: adding an array's base address to arbitrary value and its meaning

Question

I'm working on a homework assignment right now, and I'm given this information:

$s6 is the location of the base address of array A , $s0 is the location of the value of f (not specified).

It wants me to convert some instructions to a C statement. Here is my question though, because if this is answered I can very easily do the rest of this problem:

add $t0, $s6, $s0

Is that saying $t0 = (the base address of array A) + f or is it saying $t0 = A[0+f] ?

Because if the base address of A were 0x04000000 and I used addi to add 4 to that base address, yielding 0x04000004, and assigning that value to t0, what I'm saying is that t0 = A[1] (assuming I'm storing integers).

But since I don't know the value of f, I'm unsure of how to represent this in C, because I know I'm modifying the index, but I don't know by how much. Would it be more accurate to say (given the instruction above):

$t0 = A[f/4]

I'm new to all of this ha. Hopefully I demonstrated that I've done a bit of research trying to figure this out.

Thanks

OSFTW

Answer 1

There is no dereferencing in this instruction:

add $t0, $s6, $s0

It's like saying

t0 = s6 + s0;

In a C-like pseudocode. Or for your example:

t0 = (char *)A + f;

To get a value out of A, it would look something like:

lw $t1, 0($t0)

After having done the previous add instruction so that $t0 points to the right place in the array.