How does Java determine which method to call in the case of method overloading?

In method overloading, Java decides which method to execute during compilation by examining the method name and the arguments provided in the method call. The compiler searches for the most specific matching method based on the number, type, and order of parameters. This process is known as compile-time polymorphism or static binding.

Key Points:

• Method overloading is resolved at compile time, not at runtime. • The compiler selects the method with the best matching parameter list. • Method return type is not considered when resolving overloaded methods. • If multiple methods are equally suitable and the compiler cannot determine the best match, a compile-time error occurs due to ambiguity.

Example:

Suppose a Calculator class contains overloaded add() methods for int, double, and long values. When add(10, 20) is called, the compiler chooses the version that accepts int parameters because it is the closest match.

Code Example:

class Calculator {

    void display(int value) {
        System.out.println("int version");
    }

    void display(double value) {
        System.out.println("double version");
    }

    public static void main(String[] args) {

        Calculator calculator = new Calculator();

calculator.display(10); // Calls int version calculator.display(10.5); // Calls double version

    }
}

Interview Tip:

A concise interview answer is: "Java resolves method overloading at compile time by matching the method call with the most specific method signature based on the number, type, and order of arguments. If no suitable match or an ambiguous match is found, a compile-time error occurs."