R devel - Mar 2001 - lazy evaluation and DUP=F

I am having some difficulty understanding the implication of lazy evaluation
mixed
with DUP=F in a .Fortran call.  In qr.qty from base DUP is not used as an
argument so
defaults to T. I am calling qr.qty with a very large array and would like to set
DUP=F in the .Fortran call so that qr.qty would be defined as copied below. Is
there
some risk that a variable used as the argument in the original calling
environment
would be modified, or is there some other reason that this should not be done?

Also, I still find it surprising that I get very slightly different results in
some
calculations depending on whether I have DUP=T or DUP=F in a call to some of my
own
fortran. I don't think that I am reusing the arguments in anyway that would
require
duplicating them, and if I were I would expect a much bigger difference in the
result? Can anyone think of an explanation?

Thanks,
Paul Gilbert
_____

qr.qty <- function (qr, y)
{
    if (!is.qr(qr))
        stop("argument is not a QR decomposition")
    n <- as.integer(nrow(qr$qr))
    p <- as.integer(ncol(qr$qr))
    k <- as.integer(qr$rank)
    ny <- as.integer(NCOL(y))
    storage.mode(y) <- "double"
    if (NROW(y) != n)
        stop("qr and y must have the same number of rows")
    qty <- if (is.matrix(y))
        matrix(double(n * ny), n, ny)
    else double(n)
    .Fortran("dqrqty", as.double(qr$qr), n, k, as.double(qr$qraux),
        y, ny, qty = qty, PACKAGE = "base", DUP=F)$qty
}


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On Thu, 22 Mar 2001, Paul Gilbert wrote:
> I am having some difficulty understanding the implication of lazy
> evaluation mixed with DUP=F in a .Fortran call.  In qr.qty from base
> DUP is not used as an argument so defaults to T. I am calling qr.qty
> with a very large array and would like to set DUP=F in the .Fortran
> call so that qr.qty would be defined as copied below. Is there some
> risk that a variable used as the argument in the original calling
> environment would be modified, or is there some other reason that this
> should not be done?
I am not sure that DUP=FALSE can't change objects in the original calling
environment (this is why help(.Fortran) says that there is a risk). It's
not precisely lazy evaluation that causes the problem: if the lazy
evaluation really worked under the hood the way it appears to work there
would be no problem. The problem is that R actually calls by
reference until an object is modified. It should be fairly easy to tell by
experiment whether this can happen

	-thomas

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