How can one explain this seemingly inconsistent Java varargs behavior?

Question

If I write the Java method

public static void f(int... x) {
    for (int a: x) {
        System.out.println(a);
    }
}

then I can call this method via

f(1, 2, 3);

as well as

f(new int[]{1, 2, 3});

and both calls are treated exactly the same. However, the two calls

Arrays.asList(1, 2, 3)              // (a) produces a three-element Integer list

and

Arrays.asList(new int[]{1, 2, 3})   // (b) produces a one-element list of Integer arrays

are not treated the same. The section on evaluating arguments in the JLS states "The final formal parameter of m necessarily has type T[] for some T" so why isn't the "final formal parameter" in case (b) above not an integer array which would create a three-element list, like case (a)?

The only thing I can think of is that Arrays.asList is generic, and something about type compatibility (mentioned in JLS Section 15.12.4.2) is at play here, but I can't quite put my finger on the actual sentence in the JLS that makes the two cases different.

The commentary in the JLS talks about the need to craft these semantics carefully in order to handle the interplay between "parameterized types and array types that occurs in a Java Virtual Machine with erased generics" but what we end up is my function f and Arrays.asList appearing to behave inconsistent. In the former case I can "unpack" or "pack" my args and all is the same; in the latter case, the two cases are different.

How is this explained in language-lawyer terms?

I have a feeling you can reproduce the apparent inconsistency with public static void f(Object... x) or public T static void f(T... x) .. (not at a comp now so can't try) — ADTC
– ADTC, Commented Nov 26, 2014 at 12:22

ajb · Accepted Answer · 2014-11-26 06:44:26Z

5

Arrays.asList is defined as having parameter T..., and the body will see it as one argument of type T[]. However, T must be a reference type (like any type parameter), and int is not a reference type. Because of that, there's no way to make new int[]{...} compatible with an array for varargs purposes. (And the result would be List<int> which isn't allowed.) The only alternative is to treat it as a single object, and thus T here will be int[], which is a reference type, and the function will return a List<int[]>.

I think that's the reason, although I would still need to look up the exact language. It probably involves the type inference rules, which I haven't mastered.

However, I did try this:

List x = Arrays.asList(new int[]{1, 2, 3});
List y = Arrays.asList(new Integer[]{1, 2, 3});
System.out.println(x.size());   
System.out.println(y.size());

(note that I deliberately used a List raw type, something I wouldn't do in real code). The output:

1
3

edited Nov 26, 2014 at 6:44

answered Nov 26, 2014 at 6:39

ajb

31.8k4 gold badges63 silver badges86 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

TacB0sS Over a year ago

I agree, when the params are ints, each gets boxed to be a reference type, while int[] is already a reference type

NoobEditor · Accepted Answer · 2014-11-26 08:37:53Z

2

    Arrays.asList(1, 2, 3);

1 2 3 are autoboxed to Integer objects which are then placed in an Object[]

    Arrays.asList(new int[]{1, 2, 3});

forces the 3 integers into a primitive int array.

Arrays.asList() expects Object[], so it places your int[] array into a single element Object[]

edited Nov 26, 2014 at 8:37

NoobEditor

16k20 gold badges85 silver badges117 bronze badges

answered Nov 26, 2014 at 6:40

slipperyseal

2,7881 gold badge16 silver badges15 bronze badges

2 Comments

newacct Over a year ago

"Arrays.asList() expects Object[]" or T[] for any reference type T

slipperyseal Over a year ago

agreed. the winning answer explains it in better detail than me. the answer didn't exist when i wrote mine. <sobs uncontrollably>

Collectives™ on Stack Overflow

How can one explain this seemingly inconsistent Java varargs behavior?

2 Answers 2

1 Comment

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related