How is "/a/./b/../../c/" equal to /c?

I saw this as a question on one of the Stack Exchange sites. Apparently .. means to pop the stack(?). Why is this the case?


3 Answers 3


Assume root looks like:


Let's break it down to componenets:
/ -> root
/a -> in (a)
. -> THIS dir path
/a/./ -> still in /a
/a/./b -> in /a/b
.. -> go "up" one level
/a/./b/.. -> /a/b/.. -> /a
/a/./b/../.. -> /a/.. -> /
/a/./b/../../c -> /c


In the *nix world, every directory is a child directory of a parent directory. Every directory has an implicit '.' directory that refers to itself, and an implicit '..' directory that refers to its parent. So if you're in a particular directory, and you change directory (cd) to '.' (commands starting with '$' can be run on your command line!), you'll stay in the same directory:

$ pwd
$ cd .
$ pwd

But changing directory to '..' goes up one:

$ cd ..
$ pwd

And changing directory to '..' goes up again:

$ cd ..
$ pwd

The only directory that is an exception to this rule is the root directory, which doesn't have a parent directory:

$ cd ..
$ pwd
$ cd ..
$ pwd

Check out this tutorial on Unix directories for more details.


/ is your root directory. It has a directory structure like,

___ /c
___ /a
______/b (child dir of a)
___ /other dirs

Now , a single dot . means the same directory and double dots (..) Means the parent directory.

So in your example , when you use . in the path it stays in the same directory and when you use .. It jumps back to its parent directory. So eventually by jumping up, it reaches the root (/) so it's equivalent to /c

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