为什么不是我的四边形。方程类工作？

Question

我的回归价值不是正确的，而我只是想知道我做错了什么，我的代码看起来很好？！

即将出现的一些测试是：

Test failed: expected was not equal to actual

预期= 9.000000且实际= 0.000000

The discriminant is not being computed correctly

Test failed: expected was not equal to actual

预期= -3.000000且实际= 0.000000

public class QuadraticEquation {

//coefficients
private double a;
private double b;
private double c;
// created a discriminant instance variable so it's easier to access
// rather than tying out "Math.pow(b, 2) - 4 * a * c" every time
// the discriminant is required
private double discriminant = Math.pow(b, 2) - (4 * a * c);

//constructor
public QuadraticEquation(double a, double b, double c) {
    this.a = a;
    this.b = b;
    this.c = c;
}

//getter for A
public double getA() {
    return a;
}

//getter for B
public double getB() {
    return b;
}

//getter for C
public double getC() {
    return c;
}

// get discriminant
// the discriminant is inside of the square root,
//  and b^2 - 4ac is the expression for the discriminant
public double getDiscriminant() {
    //b^2 - 4ac
    return discriminant;
}

//get root0
public double getRoot0(){
    if (discriminant > 0) {
        return -b + Math.sqrt(discriminant) / 2*a;
    }
    return Double.NaN;
}

public double getRoot1() {
    if (discriminant > 0) {
        return -b - Math.sqrt(discriminant) / 2*a;
    }
    return Double.NaN;
}
}

Answer 1

您在设置a，b和c之前尝试计算判别式，因此它自然会出现零。< / p>

将计算判别式的行向下移动到构造函数中。

public class QuadraticEquation {

    //coefficients
    private double a;
    private double b;
    private double c;
    private double discriminant;

    //constructor
    public QuadraticEquation(double a, double b, double c) {
        this.a = a;
        this.b = b;
        this.c = c;
        discriminant =  b * b - (4 * a * c);
    }

    // etc.

另请注意，由于后者的计算方式，b * b通常优于Math.pow(b, 2)。

此外，getRoot0和getRoot1内的操作发生的顺序错误。你需要括号来做这样的事情。

public double getRoot0(){
    if (discriminant > 0) {
        return ( -b + Math.sqrt(discriminant)) / (2 * a);
    }
    return Double.NaN;
}

，同样在getRoot1。