Global variables to local variables - c

Question

My program to find the roots of a function using the quadratic formula is below. It works perfectly. However I was unable to get it to work without defining three of the variables globally; which, according to my project description, I am not supposed to do.
Any suggestions or alterations on how to define them locally and have the calculations not get lost before the print results function is able to do so?
#include <stdio.h>
#include <math.h>
double discriminant;
double root_one = 0, root_two = 0;
double a = 0, b = 0, c = 0;
int checkComplex(double a, double b, double c)
{
discriminant = (b * b) - 4 * (a * c);
if (discriminant == 0)
return 2;
else if (discriminant > 0)
return 1;
else
return 0;
}// end checkComplex
void calculateRoots(double a, double b, double c)
{
root_one = (-b + sqrt(discriminant)) / (2 * a);
root_two = (-b - sqrt(discriminant)) / (2 * a);
} // end calculateRoots
void getData()
{
printf("Enter a: ");
scanf("%lf", &a);
printf("\nEnter b: ");
scanf("%lf", &b);
printf("\nEnter c: ");
scanf("%lf", &c);
}// end getData
void printResults()
{
if (checkComplex(a, b, c) == 1)
{
calculateRoots(a, b, c);
printf("\n\n-----------------------------------------\n");
printf("\nThe quantity (b^2-4ac) is %.2lf", discriminant);
printf("\n\nfirst root = %.2lf\nsecond root = %.2lf\n\n", root_one,
root_two);
}// if discriminant is 1
else if (checkComplex(a, b, c) == 0)
{
printf("\n\n-----------------------------------------\n");
printf("The discriminant (b^2-4ac) is negative (imaginary)");
printf("\nTherefore, the roots are complex\n");
} // if discriminant is 0
else if (checkComplex(a == 2, b == 2, c == 2))
{
calculateRoots(a, b, c);
printf("\n\n-----------------------------------------\n");
printf("\nThe quantity (b^2-4ac) is %.2lf", discriminant);
printf("\n\nfirst root = %.2lf\nsecond root = %.2lf\n\n", root_one,
root_two);
}// if discriminant is greater than 1
} // end printResults
int main()
{
getData();
printResults();
return 0;
} // End program

Answer 1

If you need to return multiple values, you can return a struct, or you can accept pointers to where the result should be stored. I've used both approaches in the following solution:
#include <stdio.h>
#include <math.h>
typedef struct {
double discriminant;
char num_roots;
double roots[2];
} roots_t;
void getData(double* ap, double* bp, double* cp) {
printf("Enter a: "); scanf("%lf", ap);
printf("Enter b: "); scanf("%lf", bp);
printf("Enter c: "); scanf("%lf", cp);
}
roots_t calculateRoots(double a, double b, double c) {
roots_t roots;
roots.discriminant = (b*b) - 4 * (a*c);
roots.num_roots = 0;
if (roots.discriminant >= 0) {
roots.roots[roots.num_roots++] = (-b + sqrt(roots.discriminant)) / (2 * a);
if (roots.discriminant > 0)
roots.roots[roots.num_roots++] = (-b - sqrt(roots.discriminant)) / (2 * a);
}
}
return roots;
}
void printResults(double a, double b, double c, roots_t roots) {
if (roots.num_roots == 2) {
printf("The quantity (b^2-4ac) is %.2lf\n", roots.discriminant);
printf("roots = %.2lf, %.2lf\n", roots.roots[0], roots.roots[1]);
}
else if (roots.num_roots == 1) {
printf("The quantity (b^2-4ac) is %.2lf\n", roots.discriminant);
printf("roots = %.2lf\n", roots.roots[0]);
}
else {
printf("The discriminant (b^2-4ac) is negative\n");
printf("roots = <complex>\n");
}
}
int main(void) {
double a, b, c;
getData(&a, &b, &c);
printResults(a, b, c, calculateRoots(a, b, c));
return 0;
}