在Java中没有无符号字节，但是如果你想显示一个字节，你可以这样做，

int myInt = 144;

byte myByte = (byte) myInt;

char myChar = (char) (myByte & 0xFF);

System.out.println("myChar :" + Integer.toHexString(myChar));

输出:

myChar : 90

有关更多信息，请查看如何在Java中显示十六进制/字节值。

如果你有一个函数必须传递一个有符号字节，如果你传递一个无符号字节，你期望它做什么?

为什么不能使用其他数据类型?

通常情况下，您可以使用一个字节作为一个无符号字节简单或不翻译。这完全取决于如何使用。你需要澄清你打算用它做什么。

是也不是。我一直在研究这个问题。 就像我明白这个

事实上，java已经对整数-128到127进行了签名。 在java中可以通过以下方式来表示unsigned对象:

public static int toUnsignedInt(byte x) {
    return ((int) x) & 0xff;
}

如果你把-12有符号数加为无符号数，就得到244。但是你可以再次使用这个数字，它必须被移回符号，它还是-12。

如果你尝试添加244到java字节，你会得到outOfIndexException。

欢呼声……

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);
}

在Java中，原语是有符号的，这与它们在内存/传输中的表示方式无关——一个字节只有8位，是否将其解释为有符号范围取决于您。没有神奇的旗帜说“这是有符号的”或“这是没有符号的”。

由于原语是有符号的，Java编译器将阻止您为字节分配大于+127的值(或小于-128的值)。然而，没有什么可以阻止你向下转换一个int型(或short型)来实现这一点:

int i = 200; // 0000 0000 0000 0000 0000 0000 1100 1000 (200)
byte b = (byte) 200; // 1100 1000 (-56 by Java specification, 200 by convention)

/*
 * Will print a negative int -56 because upcasting byte to int does
 * so called "sign extension" which yields those bits:
 * 1111 1111 1111 1111 1111 1111 1100 1000 (-56)
 *
 * But you could still choose to interpret this as +200.
 */
System.out.println(b); // "-56"

/*
 * Will print a positive int 200 because bitwise AND with 0xFF will
 * zero all the 24 most significant bits that:
 * a) were added during upcasting to int which took place silently
 *    just before evaluating the bitwise AND operator.
 *    So the `b & 0xFF` is equivalent with `((int) b) & 0xFF`.
 * b) were set to 1s because of "sign extension" during the upcasting
 *
 * 1111 1111 1111 1111 1111 1111 1100 1000 (the int)
 * &
 * 0000 0000 0000 0000 0000 0000 1111 1111 (the 0xFF)
 * =======================================
 * 0000 0000 0000 0000 0000 0000 1100 1000 (200)
 */
System.out.println(b & 0xFF); // "200"

/*
 * You would typically do this *within* the method that expected an 
 * unsigned byte and the advantage is you apply `0xFF` only once
 * and than you use the `unsignedByte` variable in all your bitwise
 * operations.
 *
 * You could use any integer type longer than `byte` for the `unsignedByte` variable,
 * i.e. `short`, `int`, `long` and even `char`, but during bitwise operations
 * it would get casted to `int` anyway.
 */
void printUnsignedByte(byte b) {
    int unsignedByte = b & 0xFF;
    System.out.println(unsignedByte); // "200"
}

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.