java - 为什么位或操作会导致符号扩展而位却不会？

Question

我需要在 Java 中将一个字节转换为 int，但我不想进行符号扩展，所以我做了

byte b = -1
(int) (b & 0xF) // this returns 15, which is what I want
(int) (b | 0)   // this returns -1, which is essentially 0xFFFF, sign extension happens, not what I want

我认为以上两个应该给出相同的结果，但事实并非如此。我一定在位操作中遗漏了一些东西。

Answer 1

诀窍是打印这些值的二进制表示并对它们执行二进制操作

byte b = -1;
int a = (int) (b & 0xF); // this returns 15, which is what I want
int c = (int) (b | 0); // this returns -1, which is essentially 0xFFFF
System.out.println("b:" + Integer.toBinaryString(b));
System.out.println("a:" + Integer.toBinaryString(a));
System.out.println("c:" + Integer.toBinaryString(c));
System.out.println("0xF:" + Integer.toBinaryString(0xF));

打印

b:11111111111111111111111111111111
a:1111
c:11111111111111111111111111111111
0xF:1111

所以 b & OxF 是

11111111111111111111111111111111
00000000000000000000000000001111  (AND)
--------------------------------
                            1111   (15)

和b | 0 是

11111111111111111111111111111111
00000000000000000000000000000000   (OR)
--------------------------------
11111111111111111111111111111111   (-1)

Hot Licks explains why the byte value -1 is represented in binary as it is.