Bash - return a boolean check

Question

I created a simple function which does a regex check:

function is_number()
{
    return [[ "$1" =~ ^-?[0-9]+$ ]]
}

What I get is

return: [[: numeric argument required

What am I doing wrong?

Aside: The `function` keyword is a bashism that adds no significant value over the POSIX-defined function definition syntax; consider just using `is_number() {`, which will work with all POSIX-compliant shells. — Charles Duffy, May 24 '16 at 20:11
...btw, sorta-related (to the code shown as an example of contents, and including some POSIX-compliant answers): http://stackoverflow.com/questions/806906/how-do-i-test-if-a-variable-is-a-number-in-bash — Charles Duffy, May 24 '16 at 20:15

janos · Accepted Answer · 2016-05-24T19:42:54.657

6

You don't need the return statement. The return value of the function is the exit code of the last statement. So this is enough:

function is_number()
{
    [[ "$1" =~ ^-?[0-9]+$ ]]
}

The equivalent of this using an explicit return statement would be:

function is_number()
{
    [[ "$1" =~ ^-?[0-9]+$ ]]
    return $?
}

But don't do this, it's pointless.

edited May 24 '16 at 19:42

answered May 24 '16 at 19:15

janos

1 Answers1