More than one varargs argument

Question

I am getting a compile time error when I type this code out:

public void myMethod(Foo... f, Bar... b) {

}

It says that the varargs has to be the last argument, is there any other way to implement this kind of functionality?

**Any argument** after varargs is impossible, the second varags also cannot appear. — Dmitry Ginzburg, Jul 23 '14 at 09:00

Prasad Kharkar · Accepted Answer · 2014-07-23T08:55:05.333

4

This is not possible because there is no way the second argument is determined.

To understand why it should be the last argument, Consider your method

public void myMethod(Foo... f, Foo f1){}

Now suppose you call it using myMethod(fooObj1, fooObj2, fooObj3);

All the foo arguments will be applied for the var-arg method parameter. Hence there is no way to tell a specific object is passed as the second argument.

Now when you keep the var-arg as the last parameter,

public void meMethod(Foo f1, Foo... f){}

The method call myMethod(fooObj1, fooObj2, fooObj3); will assign fooObj1 to f1 and fooObj2 and fooObj3 will be applied to the var-arg f

To solve this problem, you will have to do

public void myMethod(Foo[] f, Bar... b) {

}

edited Jul 23 '14 at 08:55

answered Jul 23 '14 at 08:49

Prasad Kharkar

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Or `myMethod(Foo[] f, Bar[] b)` for consistency. – user253751 Jul 23 '14 at 09:05
@immibis obviously :) – Prasad Kharkar Jul 23 '14 at 09:05

score 0 · Answer 2 · answered Jul 23 '14 at 08:50

0

Not possible, you would have to do

public void myMethod(Foo[] f, Bar... b) {

}

as a workaround.

answered Jul 23 '14 at 08:50

CalibeR.50

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score 0 · Answer 3 · answered Jul 23 '14 at 08:51

0

You can have only one varargs in a method. Varargs must be the last formal parameter.

Try with

public void myMethod(Foo[] f, Bar... b) {

}

answered Jul 23 '14 at 08:51

bNd

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SparkOn · Answer 4 · 2014-07-23T09:03:33.020

0

Only one varargs argument is allowed in java and it must be the last parameter of the signature.Multiple vararg arguments can lead to ambiguity as you may get confused where the 1st one ends and where the second one begins.

varargs gets converted to an array, so you can simply pass two arrays instead.

edited Jul 23 '14 at 09:03

answered Jul 23 '14 at 08:55

SparkOn

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Isn't there a way for java to type check in order to determine when one type of input ends and the other starts? – Ogen Jul 23 '14 at 08:56
and how possibly can you do that? – SparkOn Jul 23 '14 at 08:57
I don't know, I was asking you. – Ogen Jul 23 '14 at 08:58

score 0 · Answer 5 · answered Jul 23 '14 at 10:16

That's one of Java's rule, in a method signature, you can have only one vararg, and at the end of the arguments list.

From the "SCJP Sun Certified Programmer for Java", here are the valid and invalid methods signatures:

// Legal:
// expects from 0 to many ints as parameters
void doStuff(int... x);
// expects first a char, then 0 to many ints
void doStuff2(char c, int... x);
// 0 to many Animals
void doStuff3(Animal... animal);

// Illegal:
// bad syntax
void doStuff4(int x...);
// too many var-args
void doStuff5(int... x, char... y);
// var-arg must be last
void doStuff6(String... s, byte b);

If you absolutely want some Foo and Bar and if they have something in common, maybe you can have method like

public void myMethod(CommonParentClassOrInterface... f);

and do some check on the actual class of each of them, but you'll have no guaranty that you'll get Foo's first and then Bar's.

I haven't really used the varargs a lot though, it's just a suggestion, but I may be wrong.

More than one varargs argument

5 Answers5