Java中的无符号字节 [英] Unsigned Bytes in Java

问题描述

Java中的字节默认是签名的。我在其他帖子上看到，使用无符号字节的解决方法类似于： int num =（int）bite& 0xFF

Bytes in Java are signed by default. I see on other posts that a workaround to have unsigned bytes is something similar to that: int num = (int) bite & 0xFF

有人可以向我解释为什么会这样做并将有符号字节转换为无符号字节然后转换为相应的整数？ ANDing 一个字节，11111111会产生相同的字节 - 对吗？

Could someone please explain to me why this works and converts a signed byte to an unsigned byte and then its respective integer? ANDing a byte with 11111111 results in the same byte - right?

推荐答案

类型转换的优先级高于& 运算符。因此，您首先要转换为int，然后进行AND运算以屏蔽所有设置的高位，包括java使用的两个补码表示法的符号位，只留下正值原始字节。例如：

A typecast has a higher precedence than the & operator. Therefore you're first casting to an int, then ANDing in order to mask out all the high-order bits that are set, including the "sign bit" of the two's complement notation which java uses, leaving you with just the positive value of the original byte. E.g.:

let byte x = 11111111 = -1
then (int) x = 11111111 11111111 11111111 11111111
and x & 0xFF = 00000000 00000000 00000000 11111111 = 255

并且您已经有效地从原始字节中删除了该符号。

and you've effectively removed the sign from the original byte.

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