在好奇netty ByteBuf writeInt和readUnsignedInt方法的明显不对称之后，我碰巧偶然地进入了这个页面。

在阅读了有趣和有教育意义的答案后，我仍然想知道你说的时候调用的是什么函数:

我试图使用这些数据作为参数的Java函数 只接受一个字节作为参数。

不管这么多年过去了，我的50美分如下:

让我们假设您正在调用的方法正在用微量更新一些余额，并且它根据一些定义良好的需求集进行操作。也就是说，它被认为对其预期的行为有正确的实现:

long processMicroPayment(byte amount) {
    this.balance += amount;
    return balance;     
}

Basically, if you supply a positive amount it will be added to the balance, and a negative amount will effectively be subtracted from the balance. Now because it accepts a byte as its parameter the implicit assumption is that it functionally only accepts amounts between -128 and +127. So if you want to use this method to add, say, 130 to the balance, it simply will not produce the result YOU desire, because there is no way within the implementation of this method to represent an amount higher than 127. So passing it 130 will not result in your desired behavior. Note that the method has no way of implementing a (say) AmountOutOfBoundsException because 130 will be 'interpreted' as a negative value that is still obeying the method's contract.

我有以下几个问题:

您是否根据其(隐式或显式)契约使用该方法? 方法是否正确实现? 我还是误解了你的问题吗?

我试图使用此数据作为参数的Java函数，只接受一个字节作为参数

这与函数接受一个大于2^32-1的整数并没有本质上的区别。

这听起来似乎取决于函数是如何定义和记录的;我认为有三种可能:

It may explicitly document that the function treats the byte as an unsigned value, in which case the function probably should do what you expect but would seem to be implemented wrong. For the integer case, the function would probably declare the parameter as an unsigned integer, but that is not possible for the byte case. It may document that the value for this argument must be greater than (or perhaps equal to) zero, in which case you are misusing the function (passing an out-of-range parameter), expecting it to do more than it was designed to do. With some level of debugging support you might expect the function to throw an exception or fail an assertion. The documentation may say nothing, in which case a negative parameter is, well, a negative parameter and whether that has any meaning depends on what the function does. If this is meaningless then perhaps the function should really be defined/documented as (2). If this is meaningful in an nonobvious manner (e.g. non-negative values are used to index into an array, and negative values are used to index back from the end of the array so -1 means the last element) the documentation should say what it means and I would expect that it isn't what you want it to do anyway.

在Java中没有原始无符号字节。通常的做法是将其转换为更大的类型:

int anUnsignedByte = (int) aSignedByte & 0xff;

Adamski提供了最好的答案，但它并不完整，所以阅读他的回复，因为它解释了我没有的细节。

如果你有一个系统函数需要传递一个无符号字节给它，你可以传递一个有符号字节，因为它会自动把它当作一个无符号字节。

因此，如果一个系统函数需要四个字节，例如，192 168 0 1作为无符号字节，您可以传递-64 -88 0 1，并且函数仍然可以工作，因为将它们传递给函数的行为将取消它们的符号。

然而，您不太可能遇到这个问题，因为系统函数隐藏在类后面以实现跨平台兼容性，尽管一些java。IO read方法返回一个int类型的未叹号字节。

如果您希望看到这种工作，请尝试将有符号字节写入文件，并将它们作为无符号字节读取回来。

我不太明白你的问题。

我刚刚尝试了这一点，对于字节-12(有符号值)，它返回整数244(相当于无符号字节值，但类型为int):

  public static int unsignedToBytes(byte b) {
    return b & 0xFF;
  }

  public static void main(String[] args) {
    System.out.println(unsignedToBytes((byte) -12));
  }

这是你想做的吗?

Java不允许像c那样将244表示为字节值。MAX_VALUE(127)你必须使用不同的整型，如short, int或long。

由于Java中的限制，无符号字节在当前的数据类型格式中几乎是不可能的。你可以为你要实现的东西使用其他语言的其他库，然后你可以使用JNI调用它们。