1

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
tshiono
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kimai 333
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  • See [Trying to call a function within a function in Bash but it returns errors](https://stackoverflow.com/q/72228033/4154375). – pjh Nov 30 '22 at 23:26

2 Answers2

1

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
knittl
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1

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.
Finally, don't use the $[…] form for arithmetic expansion. Use $((…)) instead.

M. Nejat Aydin
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