Java异常，捕获后，它仍然打印后面的内容

Question

我使用了一些异常，但即使抛出并捕获了一个异常，它也会在catch块之后继续输出whats

我希望抛出的异常被捕获，并且只打印出捕获正文中的内容，除非没有异常并转到最后一个souf

然而，不知何故，当我有一个例外，我的渔获体是打印的，但它后面的souf也不应该打印

如何组织这些异常

----引发异常的方法

   public double getHeight() throws ExceptionCheck {
        //if end point is not set, return -1 (error)
        if(points[1] == null){
            throw new ExceptionCheck("The height cannot be calculated, the end point is missing!\n\n");
        } else {
            double height = points[1].getY() - points[0].getY();
            return height;
        }
    }

------处理从getHeight抛出的方法

@Override
public double getArea() {
    //if end point is not set, return -1 (error)

    double area = 0;

    try {
        area = getHeight() * getWidth();
    }
    catch(ExceptionCheck e){
        System.out.printf("The area cannot be calculated, the end point is missing!\n\n");
    }

    return area;
}

------这里不应打印捕获后的最后一个SOUF，但无论如何都要打印

private static void printArea(Shape shape) {
    System.out.println("Printing area of a " + shape.getClass().getSimpleName());

    double area = 0d;
    // Get area of the shape and print it.
    try {
        area = shape.getArea();
    }
    catch(ExceptionCheck e){
        System.out.printf(e.getMessage());
    }
   System.out.println("The area is: " + area);
}

Answer 1

这不是catch的工作方式。如果在出现异常时不应打印，则必须将其移动到try的主体中。像

// Get area of the shape and print it.
try {
    double area = shape.getArea();
    System.out.println("The area is: " + area); // <  if the previous line throws
                                                // an exception, this will not print.
}
catch(ExceptionCheck e){
    System.out.printf(e.getMessage());
}

你的方法getArea实际上并不throw是Exception。它打印并吞下它。要调用上述catch，还必须修改getArea如下

@Override
public double getArea() throws ExceptionCheck {
    try {
        return getHeight() * getWidth();
    }
    catch(ExceptionCheck e){
        System.out.printf("The area cannot be calculated, the end point is missing!\n\n");
        throw e; // <  add this.
    }
}