value & 是什么意思?0xff 在 Java 中做什么?

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;

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

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

它将 result 设置为将 value 的 8 位放入result 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.

之所以需要这样的东西是因为 byte 在 Java 中是一个有符号类型.如果你只是写:

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 result 将以 ff ff ff fe 值结束，而不是 00 00 00 fe.更微妙的是，& 被定义为仅对 int 值¹ 进行操作，所以发生的情况是:

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 被提升为 int (ff ff ff fe).
2. 0xff 是 int 文字(00 00 00 ff).
3. 应用 & 以产生 result 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 发生在 应用 & 运算符之前.)

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

1_{嗯，不完全是.如果任一操作数是 long，& 运算符也适用于 long 值.但不在 byte 上.请参阅 Java 语言规范， 部分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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