#include <stdio.h>
#include <stdlib.h>
#include <omp.h>

/* Here are the 'function prototypes', specifying the format for all parameters. */

void AllSolve(double (*f)(double), 
                double a, double b, double L, double tol, 
                double* Roots, double* PossibleRoots,
                int sizeRoots, int* nroots, int* npossroots);

double myf(double x);

// End of function prototypes. 

int main(int argc, char* argv[])
{
    double myRoots[100];
    double myPossibleRoots[100];
    double mya=0.0;
    double myb=1.0;
    double myL = 2;
    double mytol = 1.e-3;
    int mysizeRoots = 100;
    int i, mynroots, mynpossroots;

    AllSolve(myf, mya, myb, myL, mytol,
             myRoots, myPossibleRoots,      /* No & here, because array names indicate their addresses. */
             mysizeRoots, &mynroots, &mynpossroots);

    printf("\n");    /* Here is an example of how to display the results. */
    printf("myRoots:\n");
    for (i=0; i<2*mynroots; i+=2)
    {   printf(" %f, %f \n",myRoots[i],myRoots[i+1]);
    }

    printf("\n");
    printf("myPossibleRoots:\n");
    for (i=0; i<2*mynpossroots; i+=2)
    {   printf(" %f, %f \n",myPossibleRoots[i],myPossibleRoots[i+1]);
    }
}

void AllSolve(double (*f)(double), 
                double a, double b, double L, double tol, 
                double* Roots, double* PossibleRoots,
                int sizeRoots, int* nroots, int* npossroots)
{
   double ans;
   if (sizeRoots > 3)
   {  Roots[0] = (a+b)/2.0;
      Roots[1] = Roots[0] + tol;
      Roots[2] = 3.0*(a+b)/4.0;
      Roots[3] = Roots[2] + tol;
      *nroots = 2;   /* Note that we change the contents of nroots, not
                     the pointer itself.  If we leave off the star,
                     gcc will complain about a "cast". */
                     
                     /* Note that star can mean 'contents of' or 'multiply'. */
   }
   else
   {
      *nroots = 0;
   }
   
   if (sizeRoots > 1)
   {  PossibleRoots[0] = (a+b)/4.0;
      PossibleRoots[1] = PossibleRoots[0] + tol;
      *npossroots = 1;
   }
   else
   {
      *npossroots = 0;
   }
   ans = f(a);   /* This is how the function would be called from AllSolve. */
   printf("The result of the function call in AllSolve is f(%f) = %f\n",
           a, ans);
}

double myf(double x)
{
   return  x*x - 2.0;    
}