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 that

[ -d ] && echo y

produces y is that the shell interprets it as a string in the test command and evaluates it to true. Even saying:

[ a ] && echo y

would produce y. Quoting from help test:

 string        True if string is not the null string.

---

That is why quoting variables is recommended. Saying:

[ -d "$1" ] && echo y

should not produce y when called without arguments.