lw in MIPS (also a bit C)

I have a homework question and it is troubling me. It goes like

sll $t0, $s0, 2    // $t0 = $s0 << 2;

add $t1, $t0, $s2  // $t1 = $t + $s2;

lw  $s3, $0($t1)

I'm confused about the $0, does it have the same effect as 0?

What value would the result give?

It is a question asking me to translate mips into c, where $s0 is represented as variable name a, $s1 b, $s2 c etc.

The answer for this section is supposed to be d = c[开发者_如何学Pythona];, but i really dont see why.

In MIPS, $0 or $zero is the 0th indexed and first register, and has value 0. See here.

Although that looks like a typo, since lw uses a 16-bit offset, which isn't the value from a register but rather a constant (recall that a register is 32 bits). So it should actually be lw $s3, 0($t1).

The reason the code performs d = c[a] might seem simpler if I translate the MIPS into pseudo C-code:

$t0 = a*4
$t1 = $t0 + c (= c + a*4)
d = *(c + a*4)

So we end up loading into d the value in memory at location c + 4a, which is the base address of the array c, and the index of the element we want, a. We multiply by four because the type of the array is obviously a 4-byte long type, for example a 4-byte integer, so we need to jump 4*a bytes from the beginning of the array to reach the appropriate point in memory.