Recursive function in bash

Question

I want to do a function that will return the factorial of a number in bash

Here's the current code that doesn't work, can anyone tell me what's wrong and how to correct it? I just started learning bash and I don't know that much.

#!/bash/bin factorial() {   let n=$1   if (( "$n" <= "1" ))   then return 1   else   factorial n-1   return $n*$?   fi   return 0 } factorial 5 echo "factorial 5 = $?" 

Answer 1

There are several syntax and a quite obvious logic one (return 0)

A working version is below:

#!/bin/bash  factorial() {     if (( $1 <= 1 )); then         echo 1     else         last=$(factorial $(( $1 - 1 )))         echo $(( $1 * last ))     fi } factorial 5 

You are missing:

1. return is bad (should use echo)

2. shbang line (is /bin/bash not /bash/bin)

3. Can't do arithmetic outside of (( )) or $(( )) (or let, but (( )) is preferred)

Answer 2

#!/bin/bash  function factorial()  {     if (( $1 < 2 ))    then      echo 1    else      echo $(( $1 * $(factorial $(( $1 - 1 ))) ))    fi } 

This will work better.

(It works up to 25, anyway, which should be enough to prove the point about recursion.)

For higher numbers, bc would be the tool to use, making the ninth line above:

echo "$1 * $(factorial $(( $1 - 1 )))" | bc 

but you have to be a bit careful with bc --

$ factorial 260 38301958608361692351174979856044918752795567523090969601913008174806\ 51475135399533485285838275429773913773383359294010103333339344249624\ 06009974551133984962615380298039823284896547262282019684886083204957\ 95233137023276627601257325925519566220247124751398891221069403193240\ 41688318583612166708334763727216738353107304842707002261430265483385\ 20637683911007815690066342722080690052836580858013635214371395680329\ 58941156051513954932674117091883540235576934400000000000000000000000\ 00000000000000000000000000000000000000000 

was quite a strain on my poor system!