Unix bash error - binary operator expected

Question

So below is the code i have in my bash script. I'm getting an error saying binary operator expected when i give the command 2 arguments (doesnt give error when i give 1 argument). It does change the file permissions when i give 2 arguments because i can see it when i do ls -l but it still gives me this error. How do i fix it?

for file in $@
do
    chmod 755 $file
done

if [ -z $@ ]
then
        echo "Error. No argument."
        exit $ERROR_CODE_1
fi

i have added this now

if [ ! -f "$*" ]
then
       echo "Error. File does not exist"
       exit $ERROR_NO_FILE
fi

But now when i enter more than 1 argument it just does everything in the if statement (i.e. prints error.file does not exist) even when the file does exist.

Answer 1

Doing it another way: just ask how many parameters were passed:

...
if [ $# -eq 0 ]
...

You get the error in your code because the $@ variable expands to multiple words, which leaves the test command looking like this:

[ -z parm1 parm2 parm3 ... ]