Bash if [ -d $1] returning true for empty $1

Question

So I have the following little script and keep wondering..

#!/bin/bash

if [ -d $1 ]; then
  echo \'foo\'
else
  echo \'bar\'
fi

.. why doe

Answer 1

The reason is plain and simple: The syntax does not match the case in which the -d is recognized as an operator working on a file name. It is just taken as a string, and each non-empty string is true. Only if a second parameter to -d is given, it is recognized as the operator to find out whether a given FILE is a directory.

The same applies to all the other operators like -e, -r, etc.

In your case, use double quotes to avoid running into that "problem":

[ -d "$1" ]