questions about scale by 4 for the word size from C converting to assembly language

Question

I am a beginner of assembly language, so I hope anyone who give me a answer could explain be more specific.

Question is converting from C to assembly language.

C code is:

while(save[i] == k)
    i += 1;

i and k are in $s3 and $s5 and base of array save[] is in $s6

The answer is in the figure.

What I misunderstand is that why $S3 multiple 2^2(==4), and then store into $t1.

I check on stackoverflow for similar questions, some people said "You should add the base and index together, and remember to scale by 4 for the word size. "

But because I am a beginner, I am still confused.

For example,

1. after a one cycle, i == 2 (in $S3).

2. and then start over, i == 2 in $S3 multiple by 4, we get 8 here, and then writes into $t1.

So there is a question, why is 8? I think what we need is save[2] not save[8].

I thought I might confused about value and address.

An another question is: how many bits are those register like $S1, $t1? I think is 32 bits, so it should be 1 word.

Answer 1

lw is to be provided the address of the first byte of the word to load.

If each element of save is 32 bits or 4 bytes in size, then

• save[0] is found 0 bytes beyond the start of save.
• save[1] is found 4 bytes beyond the start of save.
• save[2] is found 8 bytes beyond the start of save.
• ...

Answer 2

Each entry in save is 32 Bit long. This means, that the i-entry has an offset of i * 32 Bit = i * 4 Byte starting with the base address $s6 C is actual very smart and calculates this for you internally. So in C-Code writing save[i] is the same as *(save + i). An Addition to an pointer will be multipled by its size, so C will translate it into save + 4 * i

The Register size depends in your local architecture, but is generally speaking mostly 32 Bit for examples/exercises.