Question

所以我试图制作一个解决二次方程的C程序。我首先从头开始编写它，但它显示了相同的错误，所以我从C编程书中对它进行了一些更改。结果如下：

/*
Solves any quadratic formula.
*/
#include <stdio.h>
#include <math.h>

void main()
{
 float a, b, c, rt1= 0, rt2=0, discrim;
 clrscr();
 printf("Welcome to the Quadratic Equation Solver!");
 getch();
 printf("\nYour quadratic formula should be of the form a(x*x)+bx+c = 0");
 printf("\nPlease enter a\'s value:");
 scanf("%f", &a);
 printf("Great! Now enter b\'s value:");
 scanf("%f", &b);
 printf("One more to go! Enter c\'s value:");
 scanf("%f", &c);
 discrim = b*b - 4*a*c;
 if (discrim < 0)
  printf("\nThe roots are imaginary.");
 else
 {
  rt1 = (-b + sqrt(discrim)/(2.0*a);
  rt2 = (-b - sqrt(discrim)/(2.0*a);
  printf("\nThe roots have been calculated.");
  getch();
  printf("\nThe roots are:\nRoot 1:%f\nRoot 2:%f",rt1, rt2);
  getch();
  printf("\nThank you!");
  getch();
 }
}

Answer 1

首先，你在这里有一些不匹配的括号：

 rt1 = (-b + sqrt(discrim)/(2.0*a);
 rt2 = (-b - sqrt(discrim)/(2.0*a);

您需要将这些行更改为：

 rt1 = (-b + sqrt(discrim))/(2.0*a);
 rt2 = (-b - sqrt(discrim))/(2.0*a);
                        ^^^^

您的编译器可能会为这些行提供错误消息，例如

foo.c:25: error: expected ‘)’ before ‘;’ token

仔细查看错误消息并研究第25行会告诉您缺少右括号。

Answer 2

你错过了一些括号：

rt1 = (-b + sqrt(discrim)/(2.0*a);
rt2 = (-b - sqrt(discrim)/(2.0*a);

它应该是这样的：

rt1 = (-b + sqrt(discrim))/(2.0*a);
rt2 = (-b - sqrt(discrim))/(2.0*a);

您可以查看编译器警告。我的（g ++）打印出这样的东西：

file.c:24:36: error: expected ‘)’ before ‘;’ token

24是一个行号，您应检查错误，36是此行中的列号。

Answer 3

  rt1 = (-b + sqrt(discrim)/(2.0*a);
  rt2 = (-b - sqrt(discrim)/(2.0*a);
                          ^

你错过了正确的右括号。

Answer 4

如果您使用gcc编码，请使用以下代码并链接数学库

#include <stdio.h>
#include <math.h>

int main()
{
 float a, b, c, rt1= 0, rt2=0, discrim;
 //clrscr();
 printf("Welcome to the Quadratic Equation Solver!");
 //getch();
 printf("\nYour quadratic formula should be of the form a(x*x)+bx+c = 0");
 printf("\nPlease enter a\'s value:");
 scanf("%f", &a);
 printf("Great! Now enter b\'s value:");
 scanf("%f", &b);
 printf("One more to go! Enter c\'s value:");
 scanf("%f", &c);
 discrim = b*b - 4*a*c;
 if (discrim < 0)
  printf("\nThe roots are imaginary.");
 else
 {
  rt1 = (-b + sqrt(discrim))/(2.0*a);
  rt2 = (-b - sqrt(discrim))/(2.0*a);
  printf("\nThe roots have been calculated.");
  // getch();
  printf("\nThe roots are:\nRoot 1:%f\nRoot 2:%f",rt1, rt2);
  // getch();
  printf("\nThank you!");
  //getch();
 }
}

以这种方式编译

gcc example.c -o example -lm

二次方程式 - 不了解编译器错误

4 个答案: