Shell Script : command not found error

Question

#!/bin/bash

calc()
{
        n1=$1
        op=$2
        n2=$3
        ans=0

        if [ $# -eq 3 ]
                then
                $ans=$(expr $n1 $op $n2);
                echo "$n1 $op $n2 = $ans"
                return $ans
        else
                echo "Needs 3 parameters!"
        fi

        return;

}

I googled alot but I still can not find the error in my code, I know this is a very simple code but please help me I'm totally new and trying to self study.

The error I get is

line 12: 0=11: command not found

Thank you in advance

shellcheck helpfully points out that you shouldn't use $ on the left-hand side of an assignment. — that other guy
– that other guy, Commented Jul 25, 2016 at 20:29

bonafidegeek · Accepted Answer · 2016-07-25 20:32:02Z

4

The error is coming the '$ans' on this line

 $ans=$(expr $n1 $op $n2);

Should be

ans=$(expr $n1 $op $n2);

The '$' is evaluating the variable 'ans', as a result instead of assigning the result to your variable 'ans', it is trying to assign the result to '0'.

answered Jul 25, 2016 at 20:32

bonafidegeek

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Comments

laish138 · Accepted Answer · 2016-07-25 20:51:39Z

1

#!/bin/bash

calc()
{
        n1=$1
        op=$2
        n2=$3
        ans=0

        if [ $# -eq 3 ]
                then
                ans=$(expr $n1 $op $n2)
                echo "$n1 $op $n2 = $ans"
                return $ans
        else
                echo "Needs 3 parameters!"
        fi

        return

}

calc 6 + 5

Figured out! :)

answered Jul 25, 2016 at 20:51

laish138

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