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Dear Sietse, Thanks a lot!!!
Finally found it, after lots of trial-and-erroring through likely-looking table names in file-job.lua: the stack of input files is kept in resolvers.inputstack.
I don't program any "lua". But I was taking a look at "file-job.lua" myself. I have a lot to learn!! :-)
Assuming (absolute path of invocation dir) + (path of file relative to invocation dir) = the desired path,
I don't mind this assumption. Actually, I don't mind having a path relative to the currdir.
the following should work.
\cldcontext{ environment.runpath .. '/' .. file.pathpart(resolvers.inputstack[#resolvers.inputstack]) }
I will try this now.
You may or may not want to parse out any ./ and ../ components.
No, I don't need that.
(If you do: helpful filename manipulation functions (such as file.pathpart, indeed) may be found in l-file.lua.)
Thanks for the tip. Might be useful someday.
Hope this helps,
It does! :-) Although, I think there should be a "luatex" command for this. That is, something not related to ConTeXt. Thank you, André Caldas.