Richard Tuin Richard Tuin - 4 months ago 44
Git Question

Unable to checkout git submodule path

I have a problem when working with git submodules.

Whenever i receive a new submodule reference from the upstream repository, executing

git submodule update
gives the following result:

fatal: reference is not a tree: dd208d46ecdd1ac0d2b2594a610fe4c9150fece1
Unable to checkout 'dd208d46ecdd1ac0d2b2594a610fe4c9150fece1' in submodule path 'submodule/path'

It is important to note that the submodule has several remotes, of which the upstream remote should be used to update the submodule reference tree. I'm guessing that my problem is there, but i am not sure.

My setup is the following:

Git project


  1. origin
    (my git fork)

  2. upstream
    (project repo)

Submodule "module", has remotes:

  1. origin
    (my git fork)

  2. upstream
    (project repo)

Does anyone know what is causing my problem?


When doing git submodule update, git tries to checkout the commit/tree which is saved in the super project (in your example, the one with commit id dd208d4...)

I think you get the error because within the submodule there no such object present. You have to make sure that it is there. Usually that means that you have to fetch/pull it from a remote first.

Probably you have to

git submodule foreach git fetch
git submodule update

or maybe

git fetch --recurse-submodules

Assuming, that the submodule is configured, so that it can fetch the missing commit from the remote origin. In the end, you must know, from where you can fetch the missing commit and you have to get it.

You could check, whether you have dd208d4... by doing something like:

cd ./module
git log dd208d46ecdd1ac0d2b2594a610fe4c9150fece1
git cat-file -p dd208d46ecdd1ac0d2b2594a610fe4c9150fece1
git ls-tree dd208d46ecdd1ac0d2b2594a610fe4c9150fece1

One possible cause for such a problem is, that the one who published the new commit from the super module, did not publish the necessary commits from the submodule. He has to publish the commits from the submodule first.