[Gantz]

Assumptions

Quote

variables f,g,h,i and j are assigned to registers $s0, $s1, $s2, $s3, and $s4 respectively

The base addresses for the arrays A and B are in registers $s6 and $s7

addi $t0, $s6, 4    #load the forth element of array A into $t0
add  $t1, $s6, $0   #load the first element of array A into $t1
sw   $t1, 0($t0)    #store $t0(the forth element of array A) into $t1
lw   $t0, 0($t0)    #load $t0 into $t0
add  $s0, $t1, $t0  #add $t1 and $t0 into $s0 <=> add the forth element of array A and the first element of array A, save in $s0 

**comments are added by me and are my attempt at understanding the code**

I really don't understand lines three and four. How did these temporary registers "become arrays"?
specifically the part '0($t0)' is what's confusing me. Wouldn't this imply that $t0 is an array and we're trying to access the first element of that array?

P.S. I have to convert this code to C after, so I'll post my conversion after I've gained a better understanding of this code.

[blackcompe]

The problem is that you haven't specified whether this is an array of bytes or words (i.e. integers - 4 bytes). Since you're using the sw and lw (and not lb & sb) instructions, I'll assume it's an array of words.

addi $t0, $s6, 4   
add  $t1, $s6, $0

1. t0 = the address of A[1]
2. t1 = the address of A[0]

sw   $t1, 0($t0)    
lw   $t0, 0($t0)    

1. A[1] = address of A[0]
2. t0 = A[1]

add  $s0, $t1, $t0

1. s0 = address of A[0] + address of A[1]

This is a really weird piece of code to have someone decipher.

Quote

specifically the part '0($t0)' is what's confusing me. Wouldn't this imply that $t0 is an array and we're trying to access the first element of that array?

0($t0) is the 0th word at the address in $t0. If the base of the array is at address $t0, then 0($t0) references the first element, 4($t0) references the second, and so on. So sw $s0, 0($t0) really means A[0] = contents of s0.

[sepp2k]

Cross-posted here: http://www.daniweb.c...s-code-beginner

Duplicated work makes sepp2k sad :-(

[Gantz]

sw   $t1, 0($t0) 
lw   $t0, 0($t0)    

Wouldn't this make t1 point to A[1] and make t0 point to A[1] aswell?


So, I think that the equivalent code in C would be:

f = A[1] + A[1];

Thanks for the help blackcompe btw.

sepp2k, on 15 September 2012 - 02:58 PM, said:

Cross-posted here: http://www.daniweb.c...s-code-beginner

Duplicated work makes sepp2k sad :-(

Is that a Faux pas? I wasn't trying to offend anyone, and was only trying to access as many resources as possible -- I was really struggling with this question! (as sad as that sounds).

[sepp2k]

Gantz, on 16 September 2012 - 12:23 AM, said:

sw   $t1, 0($t0) 
lw   $t0, 0($t0)    

Wouldn't this make t1 point to A[1] and make t0 point to A[1] aswell?

The value of $t1 before those lines is the address of A[0] (or the address of A if you want). That value is not changed in the above lines because $t1 is only read on those lines - never written to.

So, no t1 will not point to A[1]. And neither will $t0 because the lw command takes the value stored in A[0] (which is the same value that is in $t1, i.e. the address of A[0], because that's what has been stored there just before) and stores it in $t0.

Quote

sepp2k, on 15 September 2012 - 02:58 PM, said:

Cross-posted here: http://www.daniweb.c...s-code-beginner

Duplicated work makes sepp2k sad :-(

Is that a Faux pas?

It leads to people answering a question without realizing that someone else already answered it on another board (as I did). Which is just an utter waste of time.

And at least on daniweb it's explicitly forbidden by the site rules. Don't know about here.

[blackcompe]

You can easily see what's happening by running this code through the MARS simulator. I ran it once just to make sure I was right, so you can rest assure what I said above is correct.