bashfunctioncommand-substitution

Bash function with parameters

I am trying to understand how to work with functions (that receive a argument) in bash. I did not get my code to work, the following code exemplifies my difficulties:

#!/bin/bash

fruit_code () {
        if [[ "$1" == "apple" ]]; then
                code=1
        else
                code=0
        fi

        return $code
}


for var in apple orange banana apple; do
        code=fruit_code $var
        echo $code
done

The shell (bash) complains saying:

apple: command not found
orange: command not found
banana: command not found
apple: command not found

So it seems the passing of parameters is not working propperly. I can not see where I am going wrong. Any help is very much appreciated. Why does it not work? What changes shoyld I do to make it work? Wish to thank you all in advance.

Kind regards Miguel

Solution

  • The first problem is that if you want to execute a command and capture its output to a variable, you need to use command substitution:

    code=$(fruit_code "$var")

    The second problem is that your function doesn't write anything to standard output; it returns an exit status. That is automatically assigned to $? after you call the function. Use this instead:

    fruit_code "$var"
echo $?

    Finally, the convention in shell is for a 0 exit status to indicate success and a non-zero value failure. In this case, your function "succeeds" if the argument is not apple.