在MIPS32程序集中复制阵列

Question

I have a segment of C++ code that I need to convert to mips using direct translation. I have most of it done, its a do while loop. I just don't understand this one line in MIPS.

x[i] = y[i];

I don't understand how to convert this into mips and I've been searching endlessly online. Could I get any help?

EDIT

This is what I had but was told it was incorrect:

la $6, y
Li $7, $1
Add $7, $7, $7
Add $7, $7, $7
Add $8, $6, $7
Lw $6, 0($8)
La $7, x
Li $8, $1
Add $8, $8, $8
Add $8, $8, $8
Add $7, $7, $8
Sw $6, 0($7)

I used this online site as reference:

http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

SECOND EDIT

C++ Coding

i=0;
do {
    x[i]=y[i];
    i++;
}
while (i!=j);

MIPS Direct translation

Addi $1, $1, 0
Loop:   la $6, y
    Li $7, $1
    Add $7, $7, $7
    Add $7, $7, $7
    Add $8, $6, $7
    Lw $6, 0($8)
    La $7, x
    Li $8, $1
    Add $8, $8, $8
    Add $8, $8, $8
    Add $7, $7, $8
    Sw $6, 0($7)
    Addi $1, $1, 1
    Bne $1, $2, loop 

And here are all my registers I can use to avoid confusion to you:

Variables   i   j   x   y   4 (constant)    Free
Registers   $1  $2   $3   $4    $5               $6, $7, $8

Answer 1

This should get you into the right direction. Since this is a homework question, I'm not going to give you a complete solution.

Quick Reference (MIPS instructions, calling convention, etc):

http://www.mips.com/media/files/MD00565-2B-MIPS32-QRC-01.01.pdf

An example assembly function that computes the dot product of a 32-bit and a 16-bit vector in a little-Endian environment:

http://code.google.com/p/mips32-asm/source/browse/dot32x16.S

Here you can see how the elements of arrays are accessed.

Note that ".set reorder" makes the assembler reorder the instructions and/or include NOPs to deal with the so-called delay slot. In case your professor wants to see that you understood the delay slot issue, you should order the instructions yourself properly and/or write your own NOP after a branch/jump.