問題描述

我有以下 Java 代碼:

I have the following Java code:

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

打印時的結果是 254，但我不知道這段代碼是如何工作的.如果 & 運算符只是按位操作，那么為什么它不會產生一個字節而是一個整數呢?

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?

推薦答案

它將 result 設置為將 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.

之所以需要這樣的東西是因為 byte 在 Java 中是一個有符號類型.如果你只是寫:

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 將以 ff ff ff fe 值結束，而不是 00 00 00 fe.更微妙的是，& 被定義為僅對 int 值¹ 進行操作，所以發生的情況是:

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 被提升為 int (ff ff ff fe).
2. 0xff 是 int 文字(00 00 00 ff).
3. 應用 & 以產生 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.

(關鍵是轉換為 int 發生在 應用 & 運算符之前.)

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

1_{嗯，不完全是.如果任一操作數是 long，& 運算符也適用于 long 值.但不在 byte 上.請參閱 Java 語言規范， 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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