Translate C++ to MIPS Assembly

Question

I need help with this last part of this problem. I'm basically trying to translate C++ code into MIPS Assembly language.

Assume a is in $s0, b in $s1, c in $s2, x in $s4, y in $s5, and z in $s6.

I finished almost all of them but I am stuck on these two, I know some parts of it but I'm having trouble putting it together as a whole. The parts I know will be followed by hashtags with the assembly code. Thanks for any help.

1.

for(x = 0; x <= z; x++) # x = 0; is: addi $s4, $0, 0
y = y + 1; # addi $s5, $s5, 1
y = 0; # addi $s5, $0, 0

2.

if(y > z)
x = 0; # addi $s4, $0, 0
else x = 1; # else: addi $s4, $0, 1

Here are the oringinal problems without the hashtags incase I am wrong:

1.

for(x = 0; x <= z; x++) 
y = y + 1; 
y = 0;

2.

if(y > z) 
x = 0; 
else x = 1;

Thanks again.

Attempt at 2, not sure if right.

ifLoop:

add $s5, ? , $s6
addi $s4, $0, 0

ifLoop

else:

addi $s4, $0, 1

else

Practice: (Assume array p is in $s7)

p[0] = 0; 
int a = 2; 
p[1] = a; 
p[a] = a;

My attempt:

sw $0, 0($s7) 
addiu $s0, $0, 2 
sw $s0, 4($s7) 
sll $t0, $s0, 2 
addu $t1, $t0, $s7 
sw $s0, 0($t1)

A for loop and an if/else would both require branch instructions (ok, that particular if/else could probably be done branchless, but I suspect that you're supposed to use branching anyway), but I see none in your translations. Look up conditional branching in your course material or an online tutorial or a MIPS instruction set reference. — Michael
Yeah, I know that I need branches in there, such as bne, beq, j, etc., however, I'm unsure of where to put them and the correct format which is why I am stuck. — Omie
Your first 3 lines are correct. I'm not familiar with sll, no idea with that line. Remember that ints take up 4 bytes, so p[a] = a in your case is p[2] = 2, so you want to eventually get to this sw $s0, 8($s7). Here's my advice: make a copy of the address located in $s7 into a temp register. Then create a counter that starts at 0. Until this counter == a, keep adding 4 to the copy of $s7. Once they are equal, you are at the address of a. — Inertiatic
@Inertiatic I tried out what you said but ended up just getting more confused lol. I'm not really understanding, sorry. If possible, do you think you can show me what you mean by writing the translated code as I don't think this book shows solutions to even numbers? And if you wan't, you can get rid of the sll since your not familiar with it, I'm just using instructions I read so far in the book. Thanks a lot and sorry for my lack of understanding. — Omie
Start a new question with this problem and comment the link here so we can put this thread to rest. Plus others might chime in. — Inertiatic

Inertiatic Inertiatic · Accepted Answer · 2014-02-20T08:25:47

Edit: 1. Luckily it is not much different without pseudo instructions.

addi $s4, $0, 0

forLoop: sle $t1, $s4, $s6  #if x <= z, set $t1 to 1, else 0
         addi $s5, $s5, 1
         addi $s5, $0, 0
         addi $s4, $s4 1
         bne $t1, $0, forLoop #repeat loop while $t1 is not 0

Here is #2. I just wanted for you to give it a go before I just gave the answer. You want to use the slt instruction to set a register to 1 or 0. If 1, the comparison is true (y > z). Then use bne to determine where to jump to. By comparing the bne to 0, the true code ends up being directly below the bne instruction. For the else, jump to the label.

slt $t2, $s6, $s5 # if z < y, set $t2 to 1, else 0
bne $t2, $0, else # if $t2==1, do the code below, if not, go to else

        addi $s4, $0, 0
        j continue    # need the jump instruction to skip the else below
else: 
        addi $s4, $0, 1

continue:
        # rest of code/program

Translate C++ to MIPS Assembly

1 Answers