Question

在下面的代码中，我试图插入方法并调用将使程序运行的代码。我试图将第一种方法放入程序中。

public class My_Geometry {

    public static void printMenu() {
        System.out.println("This is a geometry calculator, Choose what you would like to calculate: ");
        System.out.println("1) Find the area of a rectangle");
        System.out.println("2) Find the perimeter of a rectangle");
        System.out.println("3) Find the perimeter of a triangle");
    }

    public static void main(String[] args) {
        int choice;   //the user's choice
        double value;  //the value returned from the method
        char letter;  //the Y or N from the user's decision to exit
        double radius;  //the radius of the circle
        double length;  //the length of the rectangle
        double width;  //the width of the rectangle
        double height;  //the height of the triangle
        double base;  //the base of the triangle
        double side1;  //the first side of the triangle
        double side2;  //the second side of the triangle
        double side3;  //the third side of the triangle

//create a scanner object to read from the keyboard
        Scanner keyboard = new Scanner(System.in);

//do loop was chose to allow the menu to be displayed first
        letter = 'N';
        while (letter != 'Y' && letter != 'y') {

            printMenu();

            choice = keyboard.nextInt();
            value = 0.0;

            switch (choice) {
                case 1:
                    System.out.print("Enter the length of the rectangle:  ");
                    length = keyboard.nextDouble();
                    System.out.print("Enter the width of the rectangle:  ");
                    width = keyboard.nextDouble();
                    rectangleArea();
                    System.out.println("The area of the rectangle is " + value);
                    break;
                case 2:
                    System.out.print("Enter the length of the rectangle:  ");
                    length = keyboard.nextDouble();
                    System.out.print("Enter the width of the rectangle:  ");
                    width = keyboard.nextDouble();
                    // add call to rectanglePerimeter method here
                    System.out.println("The perimeter of the rectangle is " + value);
                    break;
                case 3:
                    System.out.print("Enter the length of side 1 of the triangle:  ");
                    side1 = keyboard.nextDouble();
                    System.out.print("Enter the length of side 2 of the triangle:  ");
                    side2 = keyboard.nextDouble();
                    System.out.print("Enter the length of side 3 of the triangle:  ");
                    side3 = keyboard.nextDouble();
                    // add call to trianglePerimeter method here
                    System.out.println("The perimeter of the triangle is " + value);
                    break;
                default:
                    System.out.println("You did not enter a valid choice.");
            }
            keyboard.nextLine(); //consumes the new line character after the number
            System.out.println("Do you want to exit the program (Y/N)?:  ");
            String answer = keyboard.nextLine();
            letter = answer.charAt(0);
        }
    }

}
public static void rectangleArea() {
        value = length * width;
}
// add call to rectanglePerimeter method here
// add call to trianglePerimeter method here

public static rectangleArea(){ value = length * width;} 是我试图插入程序的方法。

这是我被要求放置它的位置，但我不知道它是否写得正确。我对此方法的调用是rectangleArea()，它位于代码的中间，我相信这是调用此方法的正确方法。我只是不知道方法本身有什么问题，或者写它的正确方法是什么。

Answer 1

选项1：为返回类型声明添加void

public static void rectangleArea(){
    value = length * width;
}

并将您的变量提升到类的字段（通过移出main方法）。

选项2：将操作数移交给计算方法，让它们返回计算值并将返回值分配给main方法的局部变量：

public static double rectangleArea(double lenght, double width){
    return length * width;
}

public static void main(String[] args) {

    // ..

    double area = rectangleArea(l, w);
    System.out.println("Area: " + area);
}

Answer 2

您遇到了变量范围的问题。您已在main方法中声明了变量，因此它们仅在那里可见。使用正确的缩进格式化代码，我认为这会有所帮助。您可以考虑定义范围的每个{}。

Answer 3

rectangleArea没有返回类型。如果它什么都没有返回，那么它应该用void

声明

例如......

public static void rectangleArea(){

该方法也在类之外，在Java中是非法的。您需要在最后}

之前将其移动

value，width和length在方法中没有上下文，因为它们在main中被声明为局部变量。

您可能需要将length和width传递给该方法并返回value，例如......

public static double rectangleArea(double length, double width) {
    return length * width;
}

<强>更新...

调用该方法需要三件事。它要求您将length和width传递给方法并将返回值赋给变量（假设您对返回结果感兴趣）...

例如......

value = rectangleArea(length, width);

Answer 4

变量长度和宽度在main方法的范围内。您可以在方法外部声明长度宽度和值变量并将它们设置为静态，或者您可以将值作为方法参数传递并返回计算值。如果选择第一个建议的解决方案，则必须将void声明为方法的类型。

添加方法并调用

4 个答案: