CSSE232 : Computer Architecture I

Name: Date:

HW4 solution

In the following assembly code fragment, registers x5 and x6 are initialized to N and zero respectively.
```
LOOP: beq  x5, x0, DONE
      addi x6, x6, 1
      addi x5, x5, -1
      jal x0, LOOP
DONE:
```
a. (2 points) What is the value in register x6 for when the code reaches the label DONE:?

\(N\)

b. (2 points) How many RISC-V instructions are executed?
$$ 4N+1$$
c. (6 points) If the registers x5 and x6 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 x5, x6 and x7, translate the C code to RISC-V assembly. Use a minimum number of instructions.
```
   add x5, x0, x0
L:  bge x5, x6, done
   add x6, x6, x7
   addi x5, x5, 1
   jal x0, L
done:
```
b. (4 points) If the variables a and b are initialized to 10 and 1, what is the total number of RISC-V 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. Each iteration through the loop is four instructions, so the real number is ((2^31-1)-10) * 4 + 1 + 1 (the plus two is for the final comparison that fails).