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

[Tierney, Luke]

Thanks. Fixed in R_devel and R-patched. [STRING_IS_SORTED was not yet
used anywhere so this did not affect any computations.]

Best,

luke

On Thu, 15 Nov 2018, Michael Sannella via R-devel 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
>
> 	[[alternative HTML version deleted]]
>
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-- 
Luke Tierney
Ralph E. Wareham Professor of Mathematical Sciences
University of Iowa                  Phone:             319-335-3386
Department of Statistics and        Fax:               319-335-3017
    Actuarial Science
241 Schaeffer Hall                  email:   luke-tierney using uiowa.edu
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