Eryk,
On Sep 13, 2005, at 2:26 PM, nwew wrote:
> printf("%f\n",NUMERIC_POINTER(mat)[1]);
> [...]
> However it prints
> 0.0000
> if [EMAIL PROTECTED] are integers ( [EMAIL PROTECTED]<-matrix(1:12,3,4) ).
>
> Can anyone explain it to me why?
> I thought that NUMERIC_POINTER makes it clear that i expect
> datatype numeric.
> (Why otherwise the distinction with INTEGER_POINTER)
You answered your own question - NUMERIC_POINTER expects that the
SEXP you pass to it is numeric=double. When you use it, it's your
responsibility to make sure that the SEXP is numeric and not integer
or anything else. Probably you may want to use AS_NUMERIC to ensure
that. [btw: NUMERIC_POINTER() is a compatibility macro for REAL() and
AS_NUMERIC(x) for coerceVector(x,REALSXP)].
Also you should be aware that C uses 0-based indices so
NUMERIC_POINTER(mat)[1] accesses the 2nd element of the vector.
Cheers,
Simon
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