export proj_path

[jimhester]

For devtools I often need to construct the full paths to files inside a package, so it would be useful to have proj_path() exported so I wouldn't have to use something like file.path(usethis::proj_get() ...).

[jennybc]

The party line on that in usethis itself is that end users are supposed to use rprojroot (or here) to build paths within a project. Do you think that also applies to you (or, more generally, developers of packages that use usethis) or is the situation different? If I export proj_path(), we'd put the same caveats as with proj_get()/proj_set() about being for internal/package use, not end-user script use.

[jennybc]

The party line on that in usethis itself is that end users are supposed to use rprojroot (or here) to build paths within a project. Do you think that also applies to you (or, more generally, developers of packages that use usethis) or is the situation different? If I export proj_path(), we'd put the same caveats as with proj_get()/proj_set() about being for internal/package use, not end-user script use.

[jimhester]

I see, I guess this makes sense, you can close this if you don't think it is worthwhile.

[jimhester]

I see, I guess this makes sense, you can close this if you don't think it is worthwhile.

[jennybc]

I think it probably does make sense to export it, for use in other packages. It just comes with same risk / downsides as exporting proj_get() and proj_set().

[jennybc]

I think it probably does make sense to export it, for use in other packages. It just comes with same risk / downsides as exporting proj_get() and proj_set().