我想知道是否有人能帮我解决这个（初学者）问题。 我正在创建一个非常基本的二次方程计算器。 我在下面列出了我的代码，还有注释（代码段顶部的注释解释了我必须做的事情）。 我在网上看过多种解决方案，也尝试过自己，但似乎我一直得到不正确的x1和x2值。 如果有人能指引我，我会非常高兴。 干杯。
/*
create program to calculate x for quadratic equation. (a, b and c)
1) create function which prints out roots of quad equation.
2) throw exception if b2 - 4ac is less than 0.
3) call the function from main.
*/
#include <iostream>
#include <cmath>
#include <string>
using namespace std;
double roots() { //return type, function name, parameters and body of function.
double a = 0, b = 0, c = 0, x1, x2;
double discriminant = (b * b) - 4 * a * c;
cout << "Enter roots of quad equation (in order: a, b, c) " << endl;
cin >> a >> b >> c;
if (discriminant >= 0) {
cout << "Your quadratic equation is: " << a << "x squared + " << b << "x + " << c << " = 0 " << endl;
x1 = -b + sqrt(discriminant) / (2 * a);
x2 = -b - sqrt(discriminant) / (2 * a);
cout << "x1 = " << x1 << endl;
cout << "x2 = " << x2 << endl;
}
else {
cout << "Negative value returned from (b2 - 4ac), please try again! " << endl;
exit(1);
}
}
int main() {
cout << "Quadratic Equation Calculator " << endl;
roots();
return 0;
}`enter code here`
这是一个简单的错误。 您只是把代码行的顺序放错了。 这应该可以解决这个问题:
/*
create program to calculate x for quadratic equation. (a, b and c)
1) create function which prints out roots of quad equation.
2) throw exception if b2 - 4ac is less than 0.
3) call the function from main.
*/
#include <iostream>
#include <cmath>
#include <string>
using namespace std;
double roots() { //return type, function name, parameters and body of function.
double a = 0, b = 0, c = 0, x1, x2;
cout << "Enter roots of quad equation (in order: a, b, c) " << endl;
cin >> a >> b >> c;
double discriminant = (b * b) - 4 * a * c; // restructed this line
if (discriminant >= 0) {
cout << "Your quadratic equation is: " << a << "x squared + " << b << "x + " << c << " = 0 " << endl;
x1 = -b + sqrt(discriminant) / (2 * a);
x2 = -b - sqrt(discriminant) / (2 * a);
cout << "x1 = " << x1 << endl;
cout << "x2 = " << x2 << endl;
}
else {
cout << "Negative value returned from (b2 - 4ac), please try again! " << endl;
exit(1);
}
}
int main() {
cout << "Quadratic Equation Calculator " << endl;
roots();
return 0;
}
判别式的问题可能会教你写一个函数来计算判别式，比如:
double f_discriminant(double a,b,c){
return b*b-4*a*c;
}
这样，您的代码段就变成了（基于PranavaGande的改进）:
...
double roots() { //return type, function name, parameters and body of function.
double a = 0, b = 0, c = 0, x1, x2;
cout << "Enter roots of quad equation (in order: a, b, c) " << endl;
cin >> a >> b >> c;
double discriminant = f_discriminant(a, b, c); // restructed this line
...
这里的优点是，如果你需要再次计算判别式，你可以使用你为它编写的函数。 （对于本例，这是显而易见的，但您会遇到更难开发的函数）
问题是计算判别式的直线:
double discriminant = (b * b) - 4 * a * c;
这一行出现在您将非零值赋给a，b和C之前。 在执行时，所有这些变量都等于零，所以判别式也是零。 要使用来自用户的值计算判别式，那一行需要在这一行之后:
cin >> a >> b >> c;
原因是算术计算只执行一次，就在它所写的行上执行； 之后，=运算符（赋值）将结果放入变量中。 之后，变量就不知道值是如何到达那里的，也没有办法使用旧表达式中的新值来“刷新”自己。 你可以通过使判别式成为a，b和c的函数来近似它。
