价值与价值是什么？在Java中用0xff做什么？ [英] What does value & 0xff do in Java?

问题描述

我有以下Java代码：

I have the following Java code:

byte value = 0xfe; // corresponds to -2 (signed) and 254 (unsigned)
int result = value & 0xff;

打印时结果是254，但我不知道这段代码是如何工作的。如果& 运算符只是按位，那为什么它不会产生一个字节而是一个整数？

The result is 254 when printed, but I have no idea how this code works. If the & operator is simply bitwise, then why does it not result in a byte and instead an integer?

推荐答案

将结果设置为（无符号）值，将8位值放在在结果的最低8位。

It sets result to the (unsigned) value resulting from putting the 8 bits of value in the lowest 8 bits of result.

这样的必要原因是 byte 是Java中的签名类型。如果您刚刚写道：

The reason something like this is necessary is that byte is a signed type in Java. If you just wrote:

int result = value;

然后结果最终会得到价值 ff ff ff fe 而不是 00 00 00 fe 。另一个微妙之处是& 被定义为仅在 int 值 ¹ 上运行，所以会发生什么：

then result would end up with the value ff ff ff fe instead of 00 00 00 fe. A further subtlety is that the & is defined to operate only on int values¹, so what happens is:

1. value 升级为 int （ ff ff ff fe ）。


2. 0xff 是 int 文字（ 00 00 00 ff ）。


3. & 用于产生结果的所需值。

1. value is promoted to an int (ff ff ff fe).
2. 0xff is an int literal (00 00 00 ff).
3. The & is applied to yield the desired value for result.

（关键是转换为 int 在 <$ c $之前发生 c>& 运算符已应用。）

(The point is that conversion to int happens before the & operator is applied.)

1 _{嗯，不完全。 & 运算符也适用于 long 值，如果任一操作数为 long 。但不是 byte 。请参阅Java语言规范， 15.22.1 和 5.6.2 。}

1_{Well, not quite. The & operator works on long values as well, if either operand is a long. But not on byte. See the Java Language Specification, sections 15.22.1 and 5.6.2.}

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