Will array return by reference in Java?

Question

I found the following code in transform method of class PerspectiveTransform in JAI library:

public void transform(double[] srcPts, int srcOff,
              double[] dstPts, int dstOff,
              int numPts) {

        if ( srcPts == null ) {
            throw new IllegalArgumentException(JaiI18N.getString("Generic0"));
        }

        if (dstPts == null) {
            dstPts = new double[numPts * 2 + dstOff];
        }

        while (numPts-- > 0) {
            double x = srcPts[srcOff++];
            double y = srcPts[srcOff++];
            double w = m20 * x + m21 * y + m22;

            if (w == 0) {
                dstPts[dstOff++] = x;
                dstPts[dstOff++] = y;
            } else {
                dstPts[dstOff++] = (m00 * x + m01 * y + m02) / w;
                dstPts[dstOff++] = (m10 * x + m11 * y + m12) / w;
            }
        }
    }

Looks like if dstPts is null, then new array is created. But this array won't return outside, isn't it?

Yep, you are right, java pass by value.

nachokk
– nachokk

2013-11-14 18:56:10 +00:00
Commented Nov 14, 2013 at 18:56 — nachokk
– nachokk, Commented Nov 14, 2013 at 18:56

Fritz · Accepted Answer · 2013-11-14 19:05:14Z

3

Just a note: Java is always pass by value. The little line at the bottom is that when you pass an object as a parameter, its reference is the value that is actually passed.

For example, C# has the ref keyword, which allows the passing of a reference. In Java, you have to wrap your object in a sort of container and work with the container, so you can switch the underlying element freely.

As you say, the new array won't be accessible from the outside unless it is returned. I understand that was done to prevent the following while from throwing a NPE, but the processing result is lost as the method is actually coded.

answered Nov 14, 2013 at 19:05

Fritz

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9 Comments

PM 77-1 Over a year ago

Java is always pass by value. Exactly. No ifs or buts.

newacct Over a year ago

"is that when you pass an object as a parameter" The thing is, you cannot "pass an object as a parameter" in Java. Objects are not values in Java. The only types in Java are primitive types and reference types. Therefore, each parameter must be of a primitive type or reference type. There is no "object type".

Fritz Over a year ago

@newacct "The little line at the bottom is that when you pass an object as a parameter, its reference is the value that is actually passed." I don't know if you were clarifying or correcting me, but your quote seems out of context.

newacct Over a year ago

@Gamb: The fact that you are saying "pass an object" is itself wrong because you cannot pass something that is not a value.

Fritz Over a year ago

@newacct Again, you're missing the context. The line serves as a clarification, not an affirmation.

|

Kayaman · Accepted Answer · 2013-11-14 18:54:44Z

2

You're right, if that is the actual code. I imagine people aren't passing nulls as parameters when they're using the method, but even so that's a bug.

answered Nov 14, 2013 at 18:54

Kayaman

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Comments

Ashu · Accepted Answer · 2013-11-17 01:26:13Z

0

Arrays are Objects, and their reference is passed by value.

edited Nov 17, 2013 at 1:26

answered Nov 14, 2013 at 19:00

Ashu

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3 Comments

PM 77-1 Over a year ago

You gave a right answer but put the most important part last.

CPerkins Over a year ago

Yes. But it's a copy of the original reference. So you can't do dstPts = new double[numPts * 2 + dstOff]; in a method and expect it to have any effect outside. So you're giving a mostly correct but misleading answer.

newacct Over a year ago

Objects cannot be passed, period.

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