在方法重载中传递不同数据类型的值

Question

I pass a double value to the method that is being overloaded but it takes it as float and invokes the method for finding area of square. Float values should have an "f" to their end to be considered float right? so why does my program invoke float 'area of square' when it's supposed to invoke 'area of circle'?

public class App {
    public void calArea(float a) {
        System.out.println("Area of square is Side x Side : " + a * a);
    }

    public void calArea(float a, float b) {
        System.out.println("Area of rectangle is length x width" + a * b);
    }

    public void calArea(double r) {
        double area = (Math.PI * r * r);
        System.out.println("Area of circle is Radius x Radius x PI : " + area);
    }

    public static void main(String[] args) {

        App app = new App();

        app.calArea(5);
    }

}

Answer 1

I just post my comment as an answer, because it is the only way to resolve your problem without casting.

Instead of using double and float to distinguish what areas should be calculated. Just name them differently . This will remove a lot of headache when seeing your code.

Will app.calArea(5) calculate the area of a circle or a square?

So just change your code to this:

public class App {
    public void calAreaSquare(double a) {
        System.out.println("Area of square is Side x Side : " + a * a);
    }

    public void calAreaRectangle(double a, double b) {
        System.out.println("Area of rectangle is length x width" + a * b);
    }

    public void calAreaCircle(double r) {
        double area = (Math.PI * r * r);
        System.out.println("Area of circle is Radius x Radius x PI : " + area);
    }

    public static void main(String[] args) {

        App app = new App();

        app.calAreaCircle(5);
    }
}

I would also suggest just using double only, for the enhanced precision.

Answer 2

Simplest way to invoke method of 'area of circle' is just by calling from main method as " app.calArea(5d); ". By writing 5d or 5.0d area of circle method will be called because the value 5d is considered as double. And if you want to invoke method of 'area of square', you can do it in either way by (1) app.calArea(5f); or (2) app.calArea(5) .

So, if you want to continue with your method overloading and invoke 'area of circle' you can do it in this way

public class App {
    public void calArea(float a) {
        System.out.println("Area of square is Side x Side : " + a * a);
    }

    public void calArea(float a, float b) {
        System.out.println("Area of rectangle is length x width" + a * b);
    }

    public void calArea(double r) {
        double area = (Math.PI * r * r);
        System.out.println("Area of circle is Radius x Radius x PI : " + area);
    }

    public static void main(String[] args) {

        App app = new App();

        app.calArea(5d);
    }

}

Answer 3

Here are some quotes from the language spec that explains this behaviour.

Section 15.12.2.5 Choosing the Most Specific Method

If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run-time method dispatch. The Java programming language uses the rule that the most specific method is chosen. The informal intuition is that one method is more specific than another if any invocation handled by the first method could be passed on to the other one without a compile-time error.

In your case, both calArea(float) and calArea(double) are applicable for a single int argument ( int can be converted to either float or double through a primitive widening conversion).

Now the compiler needs to decide which is more specific. In the end, the compiler chose calArea(float) because it is more specific than calArea(double) .

Every call that is applicable to calArea(float) is applicable to calArea(double) (implicit primitive widening conversion from float to double ), but not vice versa (you need a cast to convert from double to float ).

Therefore the compiler chose the calArea(float) overload.

Answer 4

to answer "why does my program invoke float 'area of square' when it's supposed to invoke 'area of circle'?"

This is because the exact matched method which accept int argument is not found hence based on widening, nearest method that matched is called(here its float) widening happens in below order byte - > short - > int - > long - > float - > double

in your case its (int - > long - > float - > double) as you are passing int argument hence nearest match is a method that accepts float argument hence calArea(float a) is called

[to check this try including a method that accepts long argument and you will see it will get a chance to execute instead of earlier method that accepts float argument]

eg

1. public void calArea(int a) { System.out.println("int"); }

2. public void calArea(long a) { System.out.println("long"); }

3. public void calArea(float a) { System.out.println("float"); }
4. public void calArea(double a) { System.out.println("double"); }

method call is - calArea(5); case 1: lets say no method commented here method accepting int argument will be called.

case 2: now comment /remove first method (accepting int argument) here method accepting long argument will be called.

case 3: now again comment method 1 and 2 method here method accepting float argument will be called. and so on...

Answer 5

As you are passing int value not double.... so 'area of square' is being calculated. For 'area of circle' you have to pass decimal value as argument while calling the function.