bash: double vs single brackets in file test expression evaluation

Question

I have a question about file test expressions in bash. Here is a simple script to illustrate my question:

set -x
read -p "Enter a filename: " var1
if [ - e $var1 ]
then
    echo file exists
else
    echo file not found
fi

There are three scenarios:

1. At the prompt, I enter foo , which is a file that exists in the directory from which I'm running the script. As expected, the output is file exists .
2. At the prompt, I enter bar . No such file exists in the directory from which I'm running the script. As expected, the output is file not found .
3. At the prompt, I hit <enter> without typing anything. Surprisingly, the output is file exists .

If I use if [[ -e $var1 ]] , i.e., double brackets instead of single, the behavior is correct: even in the third case, I get file not found.

I stuck a set -x at the top of the file to see what was going on. With single brackets, the variable is evaluated as: '[' -e ']' . With double, it is evaluated as [[ -e '' ]] . This is interesting. Why is the expression being evaluated differently in the two cases?

I would be grateful for an explanation. Sorry if I'm missing the obvious. Thanks!

Answer 1

change to if [ -e "$var1" ]

You can find more details of [] and [[ ]] Here