when do you use varargs in java

when do you use varargs in java  using -'java,varargs'

I'm afraid of varargs. I don't know what to use them for.

Plus, it feels dangerous to let people pass as many arguments as they want.

What's an example of a context that would be a good place to use them?

asked Sep 9, 2015 by VonNeedhamjg
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8 Answers

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A good rule of thumb would be:

"Use varargs for any method (or constructor) that needs an array of T (whatever type T may be) as input".

That will make calls to these methods easier (no need to do new T[]{...}).

You could extend this rule to include methods with a List argument, provided that this argument is for input only (ie, the list is not modified by the method).

Additionally, I would refrain from using f(Object... args) because its slips towards a programming way with unclear APIs.

In terms of examples, I have used it in DesignGridLayout, where I can add several JComponents in one call:

layout.row().grid(new JLabel("Label")).add(field1, field2, field3);

In the code above the add() method is defined as add(JComponent... components).

Finally, the implementation of such methods must take care of the fact that it may be called with an empty vararg! If you want to impose at least one argument, then you have to use an ugly trick such as:

void f(T arg1, T... args) {...}

I consider this trick ugly because the implementation of the method will be less straightforward than having just T... args in its arguments list.

Hopes this helps clarifying the point about varargs.

answered Sep 9, 2015 by Kristina9175
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Varargs are useful for any method that needs to deal with an indeterminate number of objects. One good example is String.format. The format string can accept any number of parameters, so you need a mechanism to pass in any number of objects.

String.format("This is an integer: %d", myInt);
String.format("This is an integer: %d and a string: %s", myInt, myString);
answered Sep 9, 2015 by GraBowen
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Varargs can be used when we are unsure about the number of arguments to be passed in a method. It creates an array of parameters of unspecified length in the background and such a parameter can be treated as an array in runtime.

If we have a method which is overloaded to accept different number of parameters, then instead of overloading the method different times, we can simply use varargs concept.

Also when the parameters' type is going to vary then using "Object...test" will simplify the code a lot.

For example:

public int calculate(int...list) {
    int sum = 0;
    for (int item : list) {
        sum += item;
    return sum;

Here indirectly an array of int type (list) is passed as parameter and is treated as an array in the code.

For a better understanding follow this link(it helped me a lot in understanding this concept clearly): http://www.javadb.com/using-varargs-in-java

P.S: Even I was afraid of using varargs when I didn't knw abt it. But now I am used to it. As it is said: "We cling to the known, afraid of the unknown", so just use it as much as you can and you too will start liking it :)

answered Sep 9, 2015 by GladysDemwm
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I use varargs frequently for outputting to the logs for purposes of debugging.

Pretty much every class in my app has a method debugPrint():

private void debugPrint(Object... msg) {
    for (Object item : msg) System.out.print(item);

Then, within methods of the class, I have calls like the following:

debugPrint("for assignment ", hwId, ", student ", studentId, ", question ",
    serialNo, ", the grade is ", grade);

When I'm satisfied that my code is working, I comment out the code in the debugPrint() method so that the logs will not contain too much extraneous and unwanted information, but I can leave the individual calls to debugPrint() uncommented. Later, if I find a bug, I just uncomment the debugPrint() code, and all my calls to debugPrint() are reactivated.

Of course, I could just as easily eschew varargs and do the following instead:

private void debugPrint(String msg) {

debugPrint("for assignment " + hwId + ", student " + studentId + ", question "
    + serialNo + ", the grade is " + grade);

However, in this case, when I comment out the debugPrint() code, the server still has to go through the trouble of concatenating all the variables in every call to debugPrint(), even though nothing is done with the resulting string. If I use varargs, however, the server only has to put them in an array before it realizes that it doesn't need them. Lots of time is saved.

answered Sep 9, 2015 by BryanWroblew
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I have a varargs-related fear, too:

If the caller passes in an explicit array to the method (as opposed to multiple parameters), you will receive a shared reference to that array.

If you need to store this array internally, you might want to clone it first to avoid the caller being able to change it later.

 Object[] args = new Object[] { 1, 2, 3} ;

 varArgMethod(args);  // not varArgMethod(1,2,3);

 args[2] = "something else";  // this could have unexpected side-effects

While this is not really different from passing in any kind of object whose state might change later, since the array is usually (in case of a call with multiple arguments instead of an array) a fresh one created by the compiler internally that you can safely use, this is certainly unexpected behaviour.

answered Sep 9, 2015 by AndersonZFEF
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Varargs is the feature added in java version 1.5.

Why to use this?

  1. What if, you don't know the number of arguments to pass for a method?
  2. What if, you want to pass unlimited number of arguments to a method?

How this works?

It creates an array with the given arguments & passes the array to the method.

Example :

public class Solution {

    public static void main(String[] args) {

    public static void add(int... s){
        int sum=0;
        for(int num:s)
        System.out.println("sum is "+sum );


Output :


sum is 12


sum is 21

answered Sep 9, 2015 by ConcepcioTau
0 votes

I use varargs frequently for constructors that can take some sort of filter object. For example, a large part of our system based on Hadoop is based on a Mapper that handles serialization and deserialization of items to JSON, and applies a number of processors that each take an item of content and either modify and return it, or return null to reject.

answered Sep 9, 2015 by GaiMccain
0 votes

Why do we need varargs? it does look ugly to me. We maybe can use the tuples just like in Python. In that case we can not only able to accept more varibles but also return a group them without define another class for it.

answered Sep 9, 2015 by LinetteHasse