shell: why [ -d ] returns true instead of false? -

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as told here, command [ -d file ] used check whether file directory.

however, when miss file parameter in expression, still returns true.

why? how shell interpret expression then?

example should straightforward enough :)

$ [ -d /tmp ] $ echo $?          # prints 0 $ [ -d ] $ echo $?          # why prints 0 well?

it gets interpreted [ -n -d ]. in other words, since you've provided 1 argument -d gets treated if other string, , can read in the man page:

-n string
length of string nonzero

string
equivalent -n string


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