Question

我有一个像这样短的数组a：

a = [ 16748, 
      26979, 
      25888, 
      30561, 
      115
    ] //in decimal

a = [ 0100000101101100, 
      0110100101100011, 
      0110010100100000, 
      0111011101100001, 
      0000000001110011
    ] //in binary

我想通过表示每个短路的每对位来获得另一个短路阵列b。（这是难以解释的，但使用示例很容易理解）。

因此，对于数组a，我将获得数组b：

b = [ 01, 00, 00, 01, 01, 10, 11, 00, 
      01, 10, 10, 01, 01, 10, 00, 11, 
      01, 10, 01, 01, 00, 10, 00, 00, 
      01, 11, 01, 11, 01, 10, 00, 01, 
      00, 00, 00, 00, 01, 11, 00, 11
    ]

在伪代码中我想到了这样做：

int lenght = (16/2) * a.length; //16*2 because I had short (16 bit) and I want sequences of 2 bit
short[] b = new short[length]; //I create the new array of short
int j = 0; //counter of b array
foreach n in a { //foreach short in array a
    for(int i = 16 - 2; i > 0; i-2) { //shift of 2 positions to right
        b[j] = ( (n >> i) & ((2^2)-1) ); //shift and &
        j++;
    }
}

我试图将这个伪代码（将其正确）转换为Java：

public static short[] createSequencesOf2Bit(short[] a) {
    int length = (16/2) * a.length; 
    short[] b = new short[length];

    for(int i = 0; i < a.length; i++) {
        int j = 0; 
        for(short c = 16 - 2; c > 0; c -= 2) {
            short shift = (short)(a[i] >> c);
            b[j] = (short)(shift & 0x11);
            j++;
        }
    }
    return b;
}

但如果我打印b[]，我就无法获得我想要的东西。例如，仅考虑a (16748 = 0100000101101100)中的第一个短片。我得到了：

[1, 0, 16, 1, 1, 16, 17]

这是完全错误的。事实上，我应该得到：

b = [ 01, 00, 00, 01, 01, 10, 11, 00,
      ...
    ] //in binary

b = [ 1, 0, 0, 1, 1, 2, 3, 0,
      ...
    ] //in decimal

有人能帮助我吗？非常感谢。

那很奇怪。如果我只考虑a中的第一个短片而且我打印b得到：

public static short[] createSequencesOf2Bit(short[] a) {
    int length = (16/2) * a.length; 
    short[] b = new short[length];

    //for(int i = 0; i < a.length; i++) {
        int j = 0; 
        for(short c = (16 - 2); c >= 0; c -= 2) {
            short shift = (short)(a[0] >> c);
            b[j] = (short)(shift & 0x3);
            j++;
        }
    //}
    for(int i = 0; i < b.length; i++) {
        System.out.println("b[" + i + "]: " + b[i]);
    }
    return b;
} 

b = [1 0 0 1 1 2 3 0 0 0 0 0 ... 0]

但如果我打印这个：

public static short[] createSequencesOf2Bit(short[] a) {
    int length = (16/2) * a.length; 
    short[] b = new short[length];

    for(int i = 0; i < a.length; i++) {
        int j = 0; 
        for(short c = (16 - 2); c >= 0; c -= 2) {
            short shift = (short)(a[i] >> c);
            b[j] = (short)(shift & 0x3);
            j++;
        }
    }
    for(int i = 0; i < b.length; i++) {
        System.out.println("b[" + i + "]: " + b[i]);
    }
    return b;
} 

b = [0 0 0 0 1 3 0 3 0 0 0 0 ... 0]

Answer 1

我怀疑主要问题是& 0x11而不是& 0x3。0x。请注意，0x11表示十六进制，因此0b11会为您提供数字17，而不是3.或者您可以编写11来获取二进制文件中的c >= 0。

此外，正如评论中所指出的，循环条件应为c > 0，而不是{{1}}。

从短线中获得8条短裤

1 个答案: