use of the help argument in with_package

[moodymudskipper]

The help argument of with_package is not documented explicitly :

help
the name of a package, given as a name or literal character string, or a character string, depending on whether character.only is FALSE (default) or TRUE).

To understand what it does I looked into the code and saw it was just forwarded to library, there I read the same argument description in ?library but the details say :

library(help = somename) computes basic information about the package somename, and returns this in an object of class "packageInfo".

Does it make sense having this argument in with_package ?

Moreover help has no default value while other arguments that didn't have a default value in library have one in with_package. Wouldn't it be more consistent (and tidy) to have a default value for help as well (if it makes sense to keep the argument).

[jimhester]

The code currently forwards all arguments from library(), but you are correct forwarding help doesn't make a lot of sense, and confuses the issue. I think we can simply remove the argument from with_package().