使用布爾方法

Question

我很困惑我應該在哪裏以及如何使用「isSolvable」方法。我需要用它來檢查並確保分母不是0，原因很明顯。換句話說，我該如何去使用我在變量「x」和「y」上創建的「isSolvable」方法？

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; 
} 

} 

任何和所有的幫助表示讚賞。

Answer 1

大概是這樣的

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

同爲getY()。

編輯

一個例子使用這個的是趕上在使用這些方法

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