Question

What does the RISC-V code below do? Write the C equivalent. Assume that i is in register x5 and that the base address of array A that holds doubleword integers is in x20.

addi x5, x0, 0

addi x6, x0, 50

addi x28, x20, 0

loop: bge x5, x6, end

ld x7, 0(x28)

bge x7, x0, next

sub x7, x0, x7

sd x7, 0(x28)

next: addi x5, x5, 1

addi x28, x28, 8

jal x0, loop

end:

Can you rewrite the assembly code from question 4 using fewer instructions by eliminating the loop control variable? If not, why not?

Answer 1

The C code is given as :

Given x5=i and x20 = base address of A, the next element of A will be after 8 bytes because each element in A is a double word integer( 2 words = 2*4 bytes.)

Let x0 = p; x6=q; x7=r; x28=s;

#include <stdio.h>

int main()

{

i = 0;

q = 50;

s = 0;

while (i <50)

{

if( A[s] < p)

{

A[s] = p - A[s];

}

i = i + 1;

s = s + 1;

}

}

A few observations here are :

1. Given i = p + 0; and q = p + 50; and the while loop is given as i < q; we can simplify it as given in the code to avoid complexity.

2. x28 is carrying the address of next element in the array. So the term in A[s] is given by s = s+1; which is same as adding 8 bytes to the previous address.

Now coming to the next question,

This method cannot be simplified by removing loop iterator i. The loop has to run 50 times and the only way to simplify this is by expanding code 50 times which is not a good coding practice.