Bash Script - Fibonacci

Question

I was trying to make a recursive function that can calculate the Fibonacci of the enter number, by the way I got stuck on how to get the values obtained by the recursion.

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Answer 1

There are a couple of issues with your code. You cannot use $? for any useful arithmetic because the number wraps around at 255. And the embedded $((...)) is not useful at all -- you are already in arithmetic context inside the double parentheses.

#!/bin/bash

fib()
{
  local number term1 term2 # Avoid leaking to main scope
  number=$1
  if ((number < 2))
    then
      ((tmp=number))
  else
      ((--number))
      term1=$(fib "$number")
      ((--number))
      term2=$(fib "$number")
      ((tmp=term1+term2))
  fi
  ((result=result+tmp))
  printf '%s\n' "$result"
}

#Main Program.
fib "$1"   # Quote argument properly!

Inside the (( arithmetic parentheses )) you don't need the $ in front of variables; it's harmless, but you should probably try to be consistent.

As with any naïve Fibonacci implementation, this is hugely inefficient. It would be smarter to calculate the head of the sequence once in a helper function, then pull the final result and display it.

#!/bin/bash

fib2() {
    local f
    ((f=$1+$2))
    printf '%i %i\n' "$f" "$1"
}

fib()
{
    local i j
    j=$1
    shift
    for((i=1; i