Java 8 流:IntStream 到字符串字符串、Java、IntStream

2023-09-07 01:47:10 作者:爱与不爱介于一字之差

在 Java 8 流 API 中,对任何 String 对象调用 chars() 会返回一个包含所有字符的 IntStream 对象.

In Java 8 streams API, calling chars() on any String object returns an IntStream object containing all the characters.

将返回的 IntStream 对象转换回 String 的正确方法是什么?调用 toArray() 会给我一个 int[],它不被任何 String 构造函数接受.

What would be the correct way to convert the returned IntStream object back to a String? Calling toArray() would give me an int[], which is not accepted by any of the String constructor.

推荐答案

你可以使用toArray(),然后String(int[], int, int) 构造函数.这并不完全令人满意,因为 chars() 被指定返回 UTF-16 代码单元,基本上:

You can use toArray(), then the String(int[], int, int) constructor. This isn't entirely satisfactory as chars() is specified to return UTF-16 code units, basically:

返回一个 int 流,对该序列中的 char 值进行零扩展.任何映射到代理代码点的字符都会未经解释地传递.

Returns a stream of int zero-extending the char values from this sequence. Any char which maps to a surrogate code point is passed through uninterpreted.

使用 codePoints() 相反会更符合这个构造函数,它需要代码点而不是 UTF-16 代码单元.否则(使用 chars)如果您的原始字符串 确实 包含代理对,您可能会发现错误 - 我没有尝试过,但它会有意义.

Using codePoints() instead would be more in-keeping with this constructor, which expects code points rather than UTF-16 code units. Otherwise (with chars) if your original string does contain surrogate pairs, you may find you get an error - I haven't tried it, but it would make sense.

我不知道不先转换为数组的简单方法.

I don't know of a simple way of doing this without converting to an array first.