/*Translated by FOR_C, v3.4.2 (-), on 07/09/115 at 08:33:10 */
/*FOR_C Options SET: ftn=u io=c no=p op=aimnv pf=,p_dint1r s=dbov str=l x=f - prototypes */
#include <math.h>
#include "fcrt.h"
#include <stdio.h>
#include <stdlib.h>
#include "p_dint1r.h"
/*     program DRDINT1R
 *>> 1994-10-19 DRDINT1R Krogh  Changes to use M77CON
 *>> 1994-08-08 DRDINT1R Snyder  Took '0' out of formats for C conversion
 *>> 1991-11-20 CLL Edited for Fortran 90.
 *>> 1987-12-09 DRDINT1R Snyder  Initial Code.
 *--D replaces "?": DR?INT1R, ?INT1, ?INTA
 *
 *     DEMO DRIVER for 1 dimensional quadrature subprogram DINT1.
 *
 *     Compute the integral for X = 0.0 to X = PI of SIN(X), then
 *     subtract 2.0 from the ANSWER.  The result should be zero.
 *
 *     The integrand is evaluated using reverse communication.
 * */
 
int main( )
{
	long int iopt[10];
	double a, answer, b, work[1];
		/* OFFSET Vectors w/subscript range: 1 to dimension */
	long *const Iopt = &iopt[0] - 1;
	double *const Work = &work[0] - 1;
		/* end of OFFSET VECTORS */
 
 
 
	printf("\n DRDINT1R:\n Compute the integral for X = 0.0 to X = PI of SIN(X), then\n subtract 2.0 from the ANSWER.  The result should be zero.\n");
	a = 0.0e0;
	b = 4.0e0*atan( 1.0e0 );
	Iopt[2] = 10;
	Iopt[3] = 0;
	Iopt[4] = 6;
	Iopt[5] = 0;
	dint1( a, b, &answer, work, iopt );
L_30:
	;
	dinta( &answer, work, iopt );
	if (Iopt[1] == 0)
	{
		answer = sin( Work[1] );
		goto L_30;
	}
	printf("\n ANSWER          =%15.8g\n ERROR ESTIMATE  =%15.8g\n STATUS FLAG     =%6ld\n FUNCTION VALUES =%6ld\n",
	   answer, Work[1], Iopt[1], Iopt[3]);
	exit(0);
} /* end of function */