I have a parsing function that parses an encoded length from a byte buffer, it returns the parsed length as an int, and takes an index into the buffer as an integer arg. I want the function to update the index according to what it's parsed, i.e. want to pass that index by reference. In C I'd just pass an int *.
What's the cleanest way to do this in Java?
I'm currently looking at passing the index arg. as an int[], but it's a bit ugly.
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14`Integer` is immutable. – Yuval Adam Jul 24 '10 at 17:36
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1If you want to avoid the apache library, you can pass the int and return the updated value. If it's two values, I recommend using Point as a container. Otherwise, the array option or one of the other suggestions here is fine. – EntangledLoops Feb 20 '15 at 23:52
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You should be returning an int or an object; If you really need more than one, perhaps you should rethink your 'classes' design – Ujjwal Singh Sep 20 '15 at 08:41
7 Answers
You can try using org.apache.commons.lang.mutable.MutableInt from Apache Commons library. There is no direct way of doing this in the language itself.
This isn't possible in Java. As you've suggested one way is to pass an int[]. Another would be do have a little class e.g. IntHolder that wrapped an int.
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You can use java.util.concurrent.atomic.AtomicInteger.
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You cannot pass arguments by reference in Java.
What you can do is wrap your integer value in a mutable object. Using Apache Commons' MutableInt is a good option. Another, slightly more obfuscated way, is to use an int[] like you suggested. I wouldn't use it as it is unclear as to why you are wrapping an int in a single-celled array.
Note that java.lang.Integer is immutable.
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What about the usage of AtomicInteger ? Is this mutable and passed via reference ? (Afaik AtomicReference
does this correctly ). – icbytes May 23 '14 at 08:05
Wrap the byte buffer and index into a ByteBuffer object. A ByteBuffer encapsulates the concept of a buffer+position and allows you to read and write from the indexed position, which it updates as you go along.
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6Exactly. Don't force Java to do it your way, do it the Java way. Java is not C. Attempts to make it act like C will always be ugly hacks. – Skip Head Jul 24 '10 at 18:11
You can design new class like this:
public class Inte{
public int x=0;
}
later you can create object of this class :
Inte inte=new Inte();
then you can pass inte as argument where you want to pass an integer variable:
public void function(Inte inte) {
some code
}
so for update the integer value:
inte.x=value;
for getting value:
Variable=inte.x;
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You can create a Reference class to wrap primitives:
public class Ref<T>
{
public T Value;
public Ref(T value)
{
Value = value;
}
}
Then you can create functions that take a Reference as a parameters:
public class Utils
{
public static <T> void Swap(Ref<T> t1, Ref<T> t2)
{
T temp = t1.Value;
t1.Value = t2.Value;
t2.Value = temp;
}
}
Usage:
Ref<Integer> x = 2;
Ref<Integer> y = 9;
Utils.Swap(x, y);
System.out.println("x is now equal to " + x.Value + " and y is now equal to " + y.Value";
// Will print: x is now equal to 9 and y is now equal to 2
Hope this helps.
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