全局变量到局部变量

Question

下面是我使用二次公式求函数根的程序。它完美地工作。但是，如果不全局定义三个变量，我就无法让它工作；根据我的项目描述，我不应该这样做。

关于如何在本地定义它们以及在打印结果功能能够这样做之前不会丢失计算的任何建议或更改？

#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

如果你需要返回多个值，你可以返回一个结构，或者你可以接受指向应该存储结果的位置的指针。我在以下解决方案中使用了这两种方法：

#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;
}