Java Language Working with Varargs parameters


Example

Using varargs as a parameter for a method definition, it is possible to pass either an array or a sequence of arguments. If a sequence of arguments are passed, they are converted into an array automatically.

This example shows both an array and a sequence of arguments being passed into the printVarArgArray() method, and how they are treated identically in the code inside the method:

public class VarArgs {
    
    // this method will print the entire contents of the parameter passed in
    
    void printVarArgArray(int... x) {
        for (int i = 0; i < x.length; i++) {
            System.out.print(x[i] + ",");
        }
    }
    
    public static void main(String args[]) {
        VarArgs obj = new VarArgs();
        
        //Using an array:
        int[] testArray = new int[]{10, 20};
        obj.printVarArgArray(testArray); 
       
        System.out.println(" ");
        
        //Using a sequence of arguments
        obj.printVarArgArray(5, 6, 5, 8, 6, 31);
    }
}

Output:

10,20, 
5,6,5,8,6,31

If you define the method like this, it will give compile-time errors.

void method(String... a, int... b , int c){} //Compile time error (multiple varargs )

void method(int... a, String b){} //Compile time error (varargs must be the last argument