Bash Script - return actually returns something else

Question

My Original Code:

    # Factorial Using Recursion

    res=1
    fact()
    {
        x=$1
        if [ $x -le 1 ]
        then
            echo "Actual - $res"
            return `expr $res`
        else    
            #echo $x
            res=$(($res * $x))
            echo "($res)"
            fact $[$x-1]                                
        fi
    }
    fact $1
    echo "Factorial of $1 = $?"

The result (stored in $res) is certainly what I want and is also correct. But somehow when it is being returned and caught by $? it becomes erroneous.

return is not meant to return arbitrary data; it returns an exit status. To return a value, write it to standard output or put it in a global variable. (No, "proper" programming practices are not always applicable in bash.) — chepner
– chepner, Commented Oct 8, 2016 at 14:56

Tom Fenech · Accepted Answer · 2016-10-08 14:14:55Z

4

The maximum value of a return code is 255 so there's no way that you can return the factorial of most integers.

Instead, your function should output the value to standard output, e.g. using echo. Capture the output of the function like result=$(fact "$x").

If you want to output other messages while the function is executing then you can print to standard error using echo 'whatever' >&2.

res=1
fact()
{
    x=$1
    if [ "$x" -le 1 ]
    then
        echo "$res"
    else    
        res=$((res * x))
        fact "$((x - 1))"
    fi
}

I've also fixed up your syntax a little bit.

edited Oct 8, 2016 at 14:14

answered Oct 8, 2016 at 13:50

Tom Fenech

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1 Comment

AnoNym Over a year ago

thanks! i'm new to bash :P and hence its just messing with me.

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