Question

我很困惑我应该在哪里以及如何使用＆＃34; isSolvable＆＃34;方法。由于显而易见的原因，我需要使用它来检查并确保分母不为0。换句话说，我该如何使用＆＃34; isSolvable＆＃34;我在变量上创建的方法＆＃34; x＆＃34;和＆＃34; y＆＃34;？

import java.util.Scanner;

public class LinearEquations {
Scanner input = new Scanner(System.in);

private double a, b, c, d, e, f = 0;
double x, y;

public double getA () {
    System.out.println("Enter the value of a: ");
    double a = input.nextDouble();
    return a;
}

public double getB () {
    System.out.println("Enter the value of b: ");
    double b = input.nextDouble();
    return b;
}

public double getC () {
    System.out.println("Enter the value of c: ");
    double c = input.nextDouble();
    return c;
}

public double getD () {
    System.out.println("Enter the value of d: ");
    double d = input.nextDouble();
    return d;
}

public double getE () {
    System.out.println("Enter the value of e: ");
    double e = input.nextDouble();
    return e;
}

public double getF () {
    System.out.println("Enter the value of f: ");
    double f = input.nextDouble();
    return f;
}

private boolean isSolvable() { 
    if ((a * d) - (b * c) == 0){
        return false;
    }
    return true;
}

public double getX() {
    double x = ((e * d) - (b * f) / (a * d) - (b * c));
    return x;
}

public double getY() {
    double y = ((a * f) - (e * c) / (a *d) - (b * c));
    return y;
}

}

感谢任何和所有帮助。

Answer 1

可能是这样的

public double getX() {
    if (isSolvable()) {
        double x = ((e * d) - (b * f) / (a * d) - (b * c));
        return x;
    }
    else {
        throw new IllegalArgumentExceprion();
    }
}

getY()相同。

修改

使用它的一个例子是在使用这些方法的代码中捕获异常

public static void main(String[] args) { ... LinearEquations le = ...; try { double x = le.getX(); } catch (IllegalArgumentException e) { System.out.println("Equation is not solveable!"); return; } ... }

使用布尔方法

1 个答案: