阵列上的就地位反转shuffle

对于FFT函数,我需要以位反转的方式对数组内的元素进行置换或混洗.这是FFT的常见任务,因为两个大小的FFT函数的大多数功能要么以位反转的方式期望或返回它们的数据.

例如,假设数组有256个元素,我想用它的位反转模式交换每个元素.这是两个例子(二进制):

Element 00000001b should be swapped with element 10000000b
Element 00010111b should be swapped with element 11101000b

等等.

任何想法如何快速,更重要:就地？

我已经有一个执行此交换的功能.写一个并不难.由于这是DSP中常见的操作,我觉得有更聪明的方法比我非常简洁的循环更好.

有问题的语言是C,但任何语言都可以.

要使用单个传递进行交换,请在增加索引中迭代所有元素.仅当索引小于反转索引时才执行交换 - 这将跳过双重交换问题以及与相同值相反且不需要交换的回文情况(元素00000000b,10000001b,10100101b).

// Let data[256] be your element array 
for (i=0; i<256; i++)
    j = bit_reverse(i);
    if (i < j)
    {
        swap(data[i],data[j]);
    }

bit_reverse()可以使用Nathaneil的位操作技巧.bit_reverse()将被调用256次,但swap()将被调用少于128次.

一种快速的方法是交换每个相邻的单个位,然后交换2位字段等.快速的方法是:

x = (x & 0x55) << 1 | (x & 0xAA) >> 1; //swaps bits
x = (x & 0x33) << 2 | (x & 0xCC) >> 2; //swapss 2-bit fields
x = (x & 0x0F) << 4 | (x & 0xF0) >> 4;

虽然很难阅读,但如果这是需要优化的东西,你可能想要这样做.

此代码使用查找表非常快速地反转64位数字.对于您的C语言示例,我还包括32位,16位和8位数字的版本(假设int是32位).在面向对象的语言(C++,C#等)中,我会重载函数.

我目前没有方便的C编译器所以,希望我没有错过任何东西.

unsigned char ReverseBits[] = 
{
  0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0, 
  0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8, 
  0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4, 
  0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC, 
  0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2, 
  0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,
  0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6, 
  0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,
  0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,
  0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9, 
  0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,
  0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,
  0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3, 
  0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,
  0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7, 
  0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF
};


unsigned long Reverse64Bits(unsigned long number)
{    
    unsigned long result;

    result = 
        (ReverseBits[ number        & 0xff] << 56) |
        (ReverseBits[(number >>  8) & 0xff] << 48) | 
        (ReverseBits[(number >> 16) & 0xff] << 40) | 
        (ReverseBits[(number >> 24) & 0xff] << 32) | 
        (ReverseBits[(number >> 32) & 0xff] << 24) |
        (ReverseBits[(number >> 40) & 0xff] << 16) | 
        (ReverseBits[(number >> 48) & 0xff] <<  8) | 
        (ReverseBits[(number >> 56) & 0xff]);

    return result;
}

unsigned int Reverse32Bits(unsigned int number)
{
    unsigned int result;

    result = 
        (ReverseBits[ number        & 0xff] << 24) |
        (ReverseBits[(number >>  8) & 0xff] << 16) | 
        (ReverseBits[(number >> 16) & 0xff] <<  8) | 
        (ReverseBits[(number >> 24) & 0xff]);

    return result;
}

unsigned short Reverse16Bits(unsigned short number)
{
    unsigned short result;

    result = 
        (ReverseBits[ number       & 0xff] <<  8) | 
        (ReverseBits[(number >> 8) & 0xff]);

    return result;
}

unsigned char Reverse8Bits(unsigned char number)
{
    unsigned char result;

    result = (ReverseBits[number]);

    return result;
}