我试图转换一个无符号的有符号字节。问题是我接收的数据是无符号的,Java不支持无符号字节,所以当它读取数据时,它将其视为有符号的。

我尝试通过下面的解决方案转换它,我从Stack Overflow。

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

但是当它再次以字节为单位转换时,我得到了相同的带符号数据。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,所以我不能使用任何其他数据类型。我该如何解决这个问题?


当前回答

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;

其他回答

如果你认为你正在寻找这样的东西。

public static char toUnsigned(byte b) {
    return (char) (b >= 0 ? b : 256 + b);
}

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。

你还可以:

public static int unsignedToBytes(byte a)
{
    return (int) ( ( a << 24) >>> 24);
}    

解释:

假设a = (byte) 133;

在内存中,它被存储为:"1000 0101"(十六进制中的0x85)

所以它的表示可以转换为unsigned=133, signed=-123(作为2的补码)

A << 24

当向左移动24位时,结果现在是一个4字节的整数,表示为:

"10000101 00000000 00000000 00000000"(或十六进制中的"0x85000000")

然后我们有

(a << 24) >>>

它又向右移动了24位,但前导都是0。结果是:

"00000000 00000000 00000000 10000101"(或十六进制中的"0x00000085")

这就是无符号表达式它等于133。

如果你试图强制转换a = (int) a; 然后会发生的是,它保留了字节的2补表示,并将其存储为int,同时也存储为2的补:

(int)“10000101”——>“11111111 11111111 11111111 10000101”

翻译过来就是-123

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?