Java：通过引用传递int的最佳方法 [英] Java : Best way to pass int by reference

问题描述

我有一个解析函数，它解析来自字节缓冲区的编码长度，它将解析的长度作为int返回，并将索引作为整数arg作为缓冲区。我希望函数根据它的解析来更新索引，即希望通过引用传递该索引。在C中，我只传递 int * 。
在Java中用最干净的方法是什么？
我目前正在考虑传递索引arg。作为 int [] ，但它有点难看。

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.

推荐答案

您可以尝试使用Apache Commons库中的 org.apache.commons.lang.mutable.MutableInt 。在语言本身没有直接的方法。

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.

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