Recursion in bash(fibonacci sequence)-CodePudding

The main problem my code doesn't do echo every time(fibonacci sequence)

#!/bin/bash

function fib(){
        if [ $1 -le 0 ]; then
                echo 0
        elif [ $1 -eq 1 ]; then
                echo 1
        else
                echo $[`fib $[$1 - 2]`   `fib $[$1 - 1]` ]
        fi
}

fib $1

i was expecting it will do echo every time. It shows:

~/Bash$ ./fn.sh 12
144

but i need it to show like this:

~/Bash$ ./fn.sh 12
0
1
1
2
3
5
8
13
21
34
55
89
144

CodePudding user response：

Your function is consuming the output of its invocation via backticks (command substitution). Only the last output is sent to the terminal. Your function will only return the n-th number of the Fibonacci sequence.

If you want to return all the numbers of the sequence up to a certain point, you can use a loop:

for i in $(seq "$i"); do
  fib "$i"
done

CodePudding user response：

Another method might be:

#!/bin/bash

fibseq () {
    echo "$1"
    if (($3 > 0)); then fibseq "$2" $(($1   $2)) $(($3 - 1)); fi
}

if (($1 > 0)); then fibseq 0 1 "$1"; fi

Note that this is just a loop disguised as recursion. This method is much more efficient than the naive recursive version to compute Fibonacci sequences. Arguments to fibseq function: $3 serves as a counter, $1 and $2 are the last two Fibonacci numbers.
As an aside note, you can replace the $(($1 $2)) with $(bc <<< "$1 $2") if you want arbitrary-precision arithmetic.