java - 返回X的低两个字节交换的结果

Question

问题及解决方案:

/** Return the result of swapping the two lower-order bytes of X.
* For example, if X is 0x12345678, then swap(X) is 0x12347856. */
static int swapLower(int X) {
    /* Solution */
    int lower = X & 0x0000ffff;
    int upper = X & 0xffff0000;

    return upper | (0xffff & ((lower << 8) | (lower >> 8)));
}

我对如何理解该解决方案感到困惑。我尝试理解其中的逻辑，但我不明白。

而且，我一开始也不知道如何提出解决方案!

编辑:

属性:

x & 1 = x
x & 0 = 0
x | 1 = 1
x | 0 = x

1. int lower = X & 0x0000ffff = X & 0b00000000000000001111111111111111 = 0b0000000000000000x₁₅ ... x₀

2. int upper = X & 0xffff0000 = X & 0b11111111111111110000000000000000 = 0bx₃₁ ... x₁₆0000000000000000

3. 降低<< 8 = 0b0000000000000000x₁₅ ... x₀ << 8 = 0b00000000x_{15 ... x000000000}

4. 下>> 8 = 0b0000000000000000x₁₅ ... x₀>> 8 = 0bssssssss00000000x₁₅子> ... x<子>8

   • (假设X是有符号数，则s是符号位；X为正则为0，X为负则为1)

5. (下 << 8)| (下>> 8) = 0b00000000x₁₅ ... x₀00000000 | 0bssssssss00000000x₁₅ ... x₈ = 0bssssssssx₁₅ ... x₀x₁₅子> ... x₈

6. 0xffff & ((下<< 8) | (下>> 8)) = 0b000000000000000001111111111111111 & 0bssssssssx₁₅ ... x₀ x₁₅ ... x₈ = 0b00000000000000000x₇ ... x₀x₁₅ ... x₈

7. 上 | (0xffff & ((下 << 8) | (下 >> 8))) = 0bx₃₁ ... x₁₆0000000000000000 | 0b00000000000000000x₇ ... x₀x₁₅ ... x₈ = x₃₁子> ... x₁₆x₇ ... x₀x₁₅ ... x₈

Answer 1

是的，这比必要的更难理解。

我认为这样更容易理解:

int lowest2     = (X & 0x000000FF) << 8; // lowest byte moved 8 bits left
int nextLowest2 = (X & 0x0000FF00) >> 8; // next lowest byte move 8 bits right
int upper4      = (X & 0xFFFF0000);
return upper4 | lowest2 | nextLowest2;

然后当 X = 0x12345678 时:

upper4 | lowest2 | nextLowest2 = 0x12340000 | 0x00007800 | 0x00000056;

但是，当然，如果您分析您提供的解决方案，您会得到相同的结果:

如果上和下定义为:

int lower = X & 0x0000FFFF
int upper = X & 0xffff0000;

然后:

X                                                = 0x12345678
lower                                            = 0x00005678
(lower << 8)                                     = 0x00567800
(lower >> 8)                                     = 0x00000056
((lower << 8) | (lower >> 8))                    = 0x00567856
(0xFFFF & ((lower << 8) | (lower >> 8)))         = 0x00007856
upper                                            = 0x12340000
upper | (0xffff & ((lower << 8) | (lower >> 8))) = 0x12345678