Java：通过引用传递 int 的最佳方法
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Java : Best way to pass int by reference
提问by fred basset
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.
我有一个解析函数，可以解析字节缓冲区中的编码长度，它将解析后的长度作为 int 返回，并将索引作为整数 arg 传入缓冲区。我希望函数根据解析的内容更新索引，即希望通过引用传递该索引。在 C 中，我只是通过一个int *. 在 Java 中最干净的方法是什么？我目前正在考虑传递索引 arg。作为int[]，但它有点丑陋。
采纳答案by doublep
You can try using org.apache.commons.lang.mutable.MutableIntfrom Apache Commons library. There is no direct way of doing this in the language itself.
您可以尝试使用org.apache.commons.lang.mutable.MutableIntApache Commons 库。在语言本身中没有直接的方法来做到这一点。
回答by Yuval Adam
You cannot pass arguments by reference in Java.
在 Java 中不能通过引用传递参数。
What you can do is wrap your integer value in a mutable object. Using Apache Commons' MutableIntis 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 intin a single-celled array.
您可以做的是将整数值包装在一个可变对象中。使用 Apache Commons'MutableInt是一个不错的选择。另一种稍微混淆的方法是使用int[]您建议的类似方法。我不会使用它，因为不清楚为什么要将 an 包装int在单细胞数组中。
Note that java.lang.Integeris immutable.
请注意，这java.lang.Integer是不可变的。
回答by mikej
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. IntHolderthat wrapped an int.
这在 Java 中是不可能的。正如您所建议的那样，一种方法是通过int[]. 另一个是有一个小类，例如IntHolder包装一个int.
回答by John Kugelman
Wrap the byte buffer and index into a ByteBufferobject. 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.
将字节缓冲区和索引包装到一个ByteBuffer对象中。ByteBuffer 封装了缓冲区+位置的概念，并允许您从索引位置读取和写入，它会随着您的进行而更新。
回答by Anoos Sb
You can design new class like this:
您可以像这样设计新类：
public class Inte{
public int x=0;
}
later you can create object of this class :
稍后您可以创建此类的对象：
Inte inte=new Inte();
then you can pass inteas argument where you want to pass an integer variable:
然后您可以inte在要传递整数变量的地方作为参数传递：
public void function(Inte inte) {
some code
}
so for update the integer value:
所以要更新整数值：
inte.x=value;
for getting value:
获取价值：
Variable=inte.x;
回答by user3070377
You can use java.util.concurrent.atomic.AtomicInteger.
您可以使用java.util.concurrent.atomic.AtomicInteger.
回答by Mickael Bergeron Néron
You can create a Reference class to wrap primitives:
您可以创建一个 Reference 类来包装原语：
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:
然后您可以创建将引用作为参数的函数：
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:
用法：
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.
希望这可以帮助。
