BASH script mv command would not work without double-quoting variables. Why?

Question

I wrote a script today as follows:

echo "Enter a directory path"

read dir

for file in $dir/[!.]*;
    do
        f=`echo $file | sed 's/ /_/g'`
        mv "${file}" "${f}"  
    done

Initially, the mv command was written as:

mv ${file} ${f}

But that line was throwing

usage: mv [-f | -i | -n] [-v] source target
   mv [-f | -i | -n] [-v] source ... directory

I was able to use google to figure out that the variables needed to be wrapped in double quotes, but I still don't understand why doing so resolved the problem?
Thanks!

Answer 1

Quoting, in shell, prevents string-splitting and glob expansion. If you don't double-quote your variables, you have no idea how many arguments each variable may expand into after these parsing steps are run.

That is:

mv $foo $bar

...may turn into...

mv ./first word second word third word destination-file

if

foo='./first word second word third word'
bar='destination-file'

---

Similarly, consider the case where you have a filename containing a glob expression:

foo='hello * world'

In that case, your mv command would get a list of all files in the current directory.

---

...or, consider the case when an argument is empty:

foo='hello world'
bar=''

In this case, instead of (properly) getting an error that you can't have file with a 0-byte name, you would be trying to rename a file named hello to world: The $bar simply disappears.