Question

This is my program “try.sh”:

in=$*
type=(even odd)

echo -e $in " is a " ${type[is_odd $in]} " number."

is_odd()
{
    return `expr $1 % 2`
}

But if I execute “./try.sh” it gives me this error:

./try.sh: line 3: is_odd 2: syntax error in expression (error token is "2")

I want the return value of the function is_odd() to be passed as an index to the array named “type”

Please tell me how can I make it work. Thanks.

Answer 1

Rather than having is_odd return its result as its status-code, I think it’s better to print its result:

is_odd()
{
    expr $1 % 2
}

Then you can use command-substitution (`...` or $(...)) to get the result:

echo -e $in " is an " ${type[$(is_odd $in)]} " number."

Though to be honest, in this specific case I’d probably just get rid of the function and use the arithmetic expression directly — and probably adjust the quoting a bit for readability:

echo -e "$in is an ${type[in % 2]} number."

(Note that double quotes "..." do not prevent parameter substitution ${...}. Only single-quotes '...' and backslashes \ do that. Also, hat-tip to jordanm for pointing out that array indices are automatically treated as arithmetic expressions, even without expr or ((...)) or whatnot.)

That said, if you really want to return 1 for “odd” and 0 for “even”, then firstly, you should rename your function to is_even (since in Bash, 0 means “successful” or “true” and nonzero values mean “error” or “false”), and secondly, you can use a follow-on command to print its return value, and then use command-substitution. Either of these should work:

echo -e "$in is an ${type[$(is_even $in ; echo $?)]} number."

echo -e "$in is an ${type[$(is_even $in && echo 0 || echo 1)]} number."

(By the way, I’ve also changed a to an in all of the above examples: in English it’s “an odd number”, “an even number”.)