# [Rd] precision when calling a C function; presence of Fortran call

Benjamin Tyner btyner at stat.purdue.edu
Tue Dec 19 19:58:02 CET 2006

```I'm trying to figure out why the presence of a Fortran call affects the
result of a floating-point operation. I have C functions

void test1(int *n, double *f){
int outC;
double c0;

c0 = (double) *n * *f;

outC = floor(c0);
printf("when f computed by R, C says %d by itself\n",outC);
}

void test2(int *n, double *f){
extern int F77_NAME(ifloor)(double *);
int outC,outFor;
double c0;

c0 = (double) *n * *f;

outFor = F77_CALL(ifloor)(&c0);
outC = floor(c0);
printf("when f computed by R, C says %d, Fortran says %d\n",outC,outFor);
}

where the Fortran function ifloor is

integer function ifloor(x)
DOUBLE PRECISION x
ifloor=x
if(ifloor.gt.x) ifloor=ifloor-1
end

void test3(){
int outC;
double f, c0;
int n;

n = 111;
f = 40. / (double) n;

c0 = (double) n * f;
outC = floor(c0);
printf("when f computed by C, C says %d by itself\n",outC);

}

void test4(){
extern int F77_NAME(ifloor)(double *);
int outC,outFor;
double f, c0;
int n;

n = 111;
f = 40. / (double) n;

c0 = (double) n * f;
outFor = F77_CALL(ifloor)(&c0);
outC = floor(c0);
printf("when f computed by C, C says %d, Fortran says %d\n",outC,outFor);

}

For convenience, I've put all this in a package at
http://www.stat.purdue.edu/~btyner/test_0.1-1.tar.gz ; just install,
load, and run test() to see the results. On my system (linux, i686),
they are:

> library(test)
> test()
when f computed by R, C says 39 by itself
when f computed by R, C says 40, Fortran says 40
when f computed by C, C says 40 by itself
when f computed by C, C says 40, Fortran says 40

That is, with n=111 and f=40/111 passed in from R, test1 gives a value
of 39. This is not a problem; I am well aware of the limitations of
floating point. However what concerns me is that test2 gives a value of
40. It's almost as if C precision is reduced by the presence of calling
the Fortran function ifloor.  Furthermore, when n and f are computed in
C, the result is always 40. So my questions are:
1. How to explain the change from 39 to 40 seemingly due to the Fortran
call being present?
2. When Fortran is not present, why does it matter whether f is computed
by R or C?

Thanks,
Ben

```