Create link to executable in parent directory (..)

Question

Create the following directories:

parent/child

Navigate to child and create a file named child (this is an executable file in my case, not sure if that makes a difference).

I need to create two "link to executable" links in the parent. I had assumed that this would work:

ln -sf ./child ../child1
ln -sf ./child ../child2

But that creates a "link to folder" (./child) in the parent directory.

If I change it to:

ln -sf -t.. ./child child1
ln -sf -t.. ./child child2

I get an error, "ln: '../child': cannot overwrite directory".

If I do it from the parent directory (which I cannot do, this is part of a Makefile recipe):

ln -sf ./child/child ./child1
ln -sf ./child/child ./child2

It works.

Note that I cannot alter the names of any directories or files.

How do I create the links when the current directory is the child?

Answer 1

That's because the first argument, TARGET, is relative to the location of the link. In other words, you're creating links called parent/child1 and parent/child2 which both link to ./child. From the perspective of those links, ./child is the directory. You need to link to ./child/child. Or better yet, create an absolute link:

ln -sf /full/path/to/parent/child/child ../child1
ln -sf /full/path/to/parent/child/child ../child2