I encountered "unary operator expected" in a Bash script

Question

In my Bash script, I have a function to return 0 or 1 (true or false) for the later main function's condition. ``` function1 () { if [[ "${1}" =~ "^ ...some regexp... $" ]] ; then return 1 else return 0 fi } ``` Then in my main function: ``` main () { for arg in ${@} ; do if [ function1 ${arg} ] ; then ... elif [ ... ] ; then ... fi done } ``` However, when I ran this script it always gave me an error message: [: function1: unary operator expected How can I fix this?

Accepted Answer

You are making the common mistake of assuming that `[` is part of the `if` command's syntax. It is not; the syntax of `if` is simply

```
if command; then
    ... things which should happen if command's result code was 0
else
    ... things which should happen otherwise
fi
```

One of the common `command`s we use is `[` which is an alias for the command `test`. It is a simple command for comparing strings, numbers, and files. It accepts a fairly narrow combination of arguments, and tends to generate confusing and misleading error messages if you don't pass it the expected arguments. (Or rather, the error messages are adequate and helpful once you get used to it, but they are easily misunderstood if you're not used.)

In your `main` function, the call to `[` appears misplaced. You probably mean

```
if function "$arg"; then
    ...
elif ... ; then ...
```

By the way, for good measure, you should also always quote your strings. Use `"$1"` not `$1`, and `"$arg"` instead of `$arg`.

The historical reasons for `test` as a general kitchen sink of stuff the authors didn't want to make part of the syntax of `if` is one of the less attractive designs of the original Bourne shell. Bash and `zsh` offer alternatives which are less unwieldy (like the `[[` double brackets in bash, which you use in your `function1` definition), and of course, POSIX `test` is a lot more well-tempered than the original creation from Bell Labs.

As an additional clarification, your function can be simplified to just

```
function1 () {
    ! [[ "$1" =~ "^ ...some regexp... $" ]]
}
```

That is, perform the test with `[[` and reverse its result code. (The "normal" case would be to return 0 for success, but maybe you are attempting to verify that the string doesn't match?)

I encountered "unary operator expected" in a Bash script

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