Implement isLink()

[vmx]

The isLink() function was removed from the IPLD Format spec (ipld/interface-ipld-format#23), in favour of putting this functionality into js-ipld instead.

The implementation from js-ipld-dag-pb can be used.

[jonahweissman]

I'd like to tackle this, if that's okay.

[jonahweissman]

So, it looks like isLink(), as implemented in js-ipld-dag-pb, accepts a binary blob, and then uses resolve() to find the link corresponding to the given path. However, in js-ipld, the function equivalent to resolve() seems to be get(), which accepts a CID. Would it be better to try to determine the CID from the binary blob, or just change isLink() to take a CID as an argument?

It's also quite possible there's something I'm missing. I'm still trying to get a sense of how all these pieces fit together.

[jonahweissman]

To clarify my question a bit: if isLink() operates on binaryblobs, how do we know which resolver to use?

[vmx]

The more I think about it, the more i wonder if isLink() is actually a useful thing. What is its use case?

Things that came to my mind, but where there are better solutions:

• You don't know the structure of the node and you want to explore it: In this case I'd expect that you want the other contents as well, so you'd call (on the format level) resolve('/').
• You want to traverse the DAG: Then you probably know the structure of the node and you don't want to traverse any link, but a specific one. So you know that it is a a link and don't need isLink().

@jonahweissman @olizilla @mikeal @diasdavid @Stebalien Can you think of cases where you'd want to have an isLink() function?

[mikeal]

I can't think of a great case because the path resolver stops at a link and links are easy to identify as the return value, especially if we move to using CID instances.

However, I could definitely see a use for either a links method that returned all links and/or a tree option that noted which paths were links.

[vmx]

I'm closing this issue as "won't fix". Having a isLink() method is not really useful.

@jonahweissman Thanks for looking into this and getting the discussion started.