CSSE232 : Computer Architecture I

Name: Box: Date:

HW7 solution

In the following assembly code fragment, registers $t1 and $t2 are initialized to N and zero respectively.
```
LOOP: beq  $t1, $0, DONE
      addi $t2, $t2, 1
      addi $t1, $t1, -1
      j LOOP
DONE:
```
a. (2 points) What is the value in register $t2 for when the code reaches the label DONE:?

\(N\)

b. (2 points) How many MIPS instructions are executed?
$$ 4N+1$$
c. (6 points) If the registers $t1 and $t2 correspond to variables i and a respectively, write the equivalent C code for the loop.

Code to initialize a and i is optional.
```
while(i != 0) {
a++;
i--;
}
```
or
```
for(i=n; i!=0; i--)
a++;
```

Consider the following C code fragment.
```
for (i=0; i<a; i++) {
  a = a + b;
}
```
a. (4 points) Assuming that the values of i, a, and b in registers $t0, $t1 and $t2, translate the C code to MIPS assembly. Use a minimum number of instructions.
```
   add $t0, $0, $0
L: slt $t3, $t0, $t1
   beq $t3, $0,  done
   add $t1, $t1, $t2
   addi $t0, $t0, 1
   j L
done:
```
b. (4 points) If the variables a and b are initialized to 10 and 1, what is the total number of MIPS instructions executed?

\(\infty\) since a grows as fast as i.

BUT, not really! a will overflow after (2^31-1)-10 iterations of the loop because it starts at 10, then overflows when it tries to add 1 to 2^31-1: a becomes negative. The next time through the loop, the comparison will fail and the loop will end. So the real number is ((2^31-1)-10) * 5 + 1 + 2 (the plus two is for the final comparison that fails).