为什么ByteArrayOutputStream使用int？ [英] Why does ByteArrayOutputStream use int?

问题描述

也许有人可以帮助我理解，因为我觉得我错过的东西可能会对我的程序运行方式产生影响。

Maybe someone can help me understand because I feel I'm missing something that will likely have an effect on how my program runs.

我正在使用ByteArrayOutputStream 。除非我错过了一些巨大的东西，否则这个类的重点是创建一个byte []数组用于其他用途。

I'm using a ByteArrayOutputStream. Unless I've missed something huge, the point of this class is to create a byte[] array for some other use.

然而，普通写入函数BAOS取一个int而不是一个字节（ ByteArrayOutputStream.write ）。

However, the "plain" write function on BAOS takes an int not a byte (ByteArrayOutputStream.write).

根据这个（原始数据类型）页面，在Java中，int是32位数据类型，字节是8位数据类型。

According to this(Primitive Data Types) page, in Java, an int is a 32-bit data type and a byte is an 8-bit data type.

如果我写这段代码......

If I write this code...

int i = 32;
byte b = i;

我收到有关可能需要更改的有损转换的警告......

I get a warning about possible lossy conversions requiring a change to this...

int i = 32;
byte b = (byte)i;

我对write（int）感到困惑......

I'm really confused about write(int)...

推荐答案

ByteArrayOutputStream 只是覆盖在 OutputStream 。所以真正的问题是为什么 OutputStream.write（int） 是这样声明的，当它声明的目标是将单个字节写入流时。流的实现在这里是无关紧要的。

ByteArrayOutputStream is just overriding the abstract method declared in OutputStream. So the real question is why OutputStream.write(int) is declared that way, when its stated goal is to write a single byte to the stream. The implementation of the stream is irrelevant here.

你的直觉是正确的 - 在我看来，这是一个破碎的设计。是的，它会丢失数据，正如文档中明确指出的那样：

Your intuition is correct - it's a broken bit of design, in my view. And yes, it will lose data, as is explicitly called out in the docs:

要写入的字节是八个低阶论点的位数b。 b的24个高位被忽略。

The byte to be written is the eight low-order bits of the argument b. The 24 high-order bits of b are ignored.

对于这个来说，这将是更明智的（在我看来） 写（字节）。唯一的缺点是你不能在没有强制转换的情况下用文字值调用它：

It would have been much more sensible (in my view) for this to be write(byte). The only downside is that you couldn't then call it with literal values without casting:

// Write a single byte 0. Works with current code, wouldn't work if the parameter
// were byte.
stream.write(0);

看起来没问题，但不是 - 因为文字0的类型是 int ，它不能隐式转换为 byte 。你必须使用：

That looks okay, but isn't - because the type of the literal 0 is int, which isn't implicitly convertible to byte. You'd have to use:

// Ugly, but would have been okay with write(byte).
stream.write((byte) 0);

对我而言，设计API的方式不是很好，但这就是我们的意思自从Java 1.0以来，它已经拥有了。不幸的是，如果没有它在整个地方发生重大变化，它现在无法修复。

For me that's not a good enough reason to design the API the way it is, but that's what we've got - and have had since Java 1.0. It can't be fixed now without it being a breaking change all over the place, unfortunately.

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