The following code is used to derive Fibonacci algorithm with this sequence: (0, 1, 1, 2, 3, 5, 8, 13, 21, …) where each number is “add” of two previous ones. int fib(int n){ if (n==0) return 0; else if (n == 1) return 1; else return fib(n−1) + fib(n−2); } Explain why the following RISC-V assembly code works, you need to explain all the details using the comments as hints? # IMPORTANT! Stack pointer must remain a multiple of 16!!!! addi x10,x10,8 # fib(8), you can change it fib: beq x10, x0, done # If n==0, return 0 addi x5, x0, 1 beq x10, x5, done # If n==1, return 1 addi x2, x2, -16 # Allocate 2 words of stack space sd x1, 0(x2) # Save the return address sd x10, 8(x2) # Save the current n addi x10, x10, -1 # x10 = n-1 jal x1, fib # fib(n-1) ld x5, 8(x2) # Load old n from the stack sd x10, 8(x2) # Push fib(n-1) onto the stack addi x10, x5, -2 # x10 = n-2 jal x1, fib # Call fib(n-2) ld x5, 8(x2) # x5 = fib(n-1) add x10, x10, x5 # x10 = fib(n-1)+fib(n-2) # Clean up: ld x1, 0(x2) # Load saved return address addi x2, x2, 16 # Pop two words from the stack done: jalr x0, 0(x1)

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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The following code is used to derive Fibonacci algorithm with this sequence: (0, 1, 1, 2, 3, 5, 8, 13, 21, …) where each number is “add” of two previous ones.

int fib(int n){

if (n==0)

return 0;

else if (n == 1)

return 1;

else

return fib(n−1) + fib(n−2);

}

Explain why the following RISC-V assembly code works, you need to explain all the details using the comments as hints?

# IMPORTANT! Stack pointer must remain a multiple of 16!!!!

addi x10,x10,8 # fib(8), you can change it
fib:
beq x10, x0, done # If n==0, return 0
addi x5, x0, 1
beq x10, x5, done # If n==1, return 1
addi x2, x2, -16 # Allocate 2 words of stack space
sd x1, 0(x2) # Save the return address
sd x10, 8(x2) # Save the current n
addi x10, x10, -1 # x10 = n-1

jal x1, fib # fib(n-1)
ld x5, 8(x2) # Load old n from the stack
sd x10, 8(x2) # Push fib(n-1) onto the stack
addi x10, x5, -2 # x10 = n-2
jal x1, fib # Call fib(n-2)
ld x5, 8(x2) # x5 = fib(n-1)
add x10, x10, x5 # x10 = fib(n-1)+fib(n-2)

# Clean up:
ld x1, 0(x2) # Load saved return address
addi x2, x2, 16 # Pop two words from the stack

done:
jalr x0, 0(x1)

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