[Rd] STRING_IS_SORTED claims as.character(1:100) is sorted

[Gabriel Becker]

Thank you for the report. We will look at this and make sure it gets fixed.

~G


On Thu, Nov 15, 2018, 3:13 PM Michael Sannella via R-devel <
r-devel using r-project.org> wrote:

> If I have loaded the C code:
>     SEXP altrep_STRING_IS_SORTED(SEXP x)
>     {
>         return ScalarInteger(STRING_IS_SORTED(x));
>     }
> and defined the function:
>     issort <- function(x) .Call("altrep_STRING_IS_SORTED",x)
>
> I am seeing the following results in R 3.5.1/Linux:
>     > issort(LETTERS)
>     [1] NA
>     > issort(as.character(1:100))  ## should return NA
>     [1] 1
>     > issort(as.character(100:1))  ## should return NA
>     [1] -1
>     > issort(as.character(1:100+1L))
>     [1] NA
>
> issort(as.character(1:100)) should return NA, since the string vector
> "1","2",..."10",... is not sorted.  I suspect that the problem is that
> the Is_sorted method for deferred_string is just calling the Is_sorted
> method for the source object 1:100 (which _is_ a sorted integer
> vector).  It should probably just return NA for any source object.
>
>   ~~ Michael Sannella
>
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>
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