How to Determine If a Java Function Was Called with Varargs or an Array

How can you determine whether a Java method was called using varargs or an array as an argument?

public class VarargsExample {
    public static void main(String[] args) {
        method(1, 2, 3); // Called with varargs
        method(new int[]{1, 2, 3}); // Called with an array
    }

    public static void method(int... numbers) {
        if (numbers instanceof int[]) {
            System.out.println("Called with varargs");
        }
    }

    public static void method(int[] numbers) {
        System.out.println("Called with an array");
    }
}

In Java, you cannot directly determine at runtime whether a method was called using varargs (variable-length argument list) or an array because both are translated to an array of objects under the hood. However, there are strategies to distinguish between the two scenarios during implementation.

public class VarargsExample {
    public static void main(String[] args) {
        method(1, 2, 3); // Varargs
        method(new int[]{1, 2, 3}); // Array
    }

    public static void method(int... numbers) {
        System.out.println("Called with varargs");
    }

    public static void method(int[] numbers) {
        System.out.println("Called with an array");
    }
}

Causes

• Java treats varargs and arrays similarly when invoked, leading to challenges in distinguishing between them directly.
• Varargs allow for a flexible number of parameters, while arrays are fixed in size and type.

Solutions

• Overload the method: Define two separate methods for handling varargs and arrays. This way, you can clearly separate the behavior for each.
• Inside the varargs method, you can check the instance type of the argument to differentiate if needed.

Mistake: Confusing varargs with a single array parameter leading to incorrect method overload resolution.

Solution: Ensure that you adequately overload methods to cater separately for varargs and arrays.

Mistake: Assuming runtime reflection can distinguish between varargs and arrays.

Solution: Understand that Java treats varargs as arrays, and design your methods accordingly.