如何用C语言实现带有字节元素数组的128位线性反馈移位寄存器

发布于 2024-12-10 18:36:48 字数 764 浏览 0 评论 0原文

我有一个如下数组，

unsigned char A[16]

我使用这个数组来表示 128 位硬件寄存器。现在我想使用这个长寄存器实现一个线性反馈移位寄存器（LFSR，斐波那契实现）。连接到该 LFSR 反馈同或门的多项式（或抽头）为 [128, 29, 27, 2, 1]。

16 位 LFSR 的实现（抽头位于 [16, 14, 13, 11]）可以从获得维基百科如下。

  unsigned short lfsr = 0xACE1u;
  unsigned bit;

  unsigned rand()
  {
    bit  = ((lfsr >> 0) ^ (lfsr >> 2) ^ (lfsr >> 3) ^ (lfsr >> 5) ) & 1;
    return lfsr =  (lfsr >> 1) | (bit << 15);
  }

然而，就我而言，我需要将位从一个字节元素转移到另一个字节元素，例如，需要将 msb 或 A[0] 转移到 A1。完成此转变的最少编码是什么？谢谢你！

原文

I have an array as follows,

unsigned char A[16]

I am using this array to represent a 128-bit hardware register. Now I want to implement a linear feedback shift register (LFSR, Fibonacci implementation) using this long register. The polynomials (or tap) which connect to the feedback xnor gate of this LFSR are [128, 29, 27, 2, 1].

The implementation of a 16-bit LFSR (taps at [16, 14, 13, 11]) can be obtained from Wikipedia as the following.

  unsigned short lfsr = 0xACE1u;
  unsigned bit;

  unsigned rand()
  {
    bit  = ((lfsr >> 0) ^ (lfsr >> 2) ^ (lfsr >> 3) ^ (lfsr >> 5) ) & 1;
    return lfsr =  (lfsr >> 1) | (bit << 15);
  }

In my case, however, I need to shift bits from one byte element to another, e.g. the msb or A[0] need to be shift to the lsb of A1. What is minimum coding to do this shift?
Thank you!

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多彩岁月 2024-12-17 18:36:48

要计算要移入的位，您不需要每次都移动整个数组，因为您只对一位感兴趣（请注意 bit = 末尾的 & 1 行来自维基百科）。

正确的移位量是：

128 - 128 =   0   => byte  0 bit 0
128 -  29 =  99   => byte 12 bit 3
128 -  27 = 101   => byte 12 bit 5
128 -   2 = 126   => byte 15 bit 6
128 -   1 = 127   => byte 15 bit 7

所以，

bit = ((A[0] >> 0) 
    ^  (A[12] >> 3) 
    ^  (A[12] >> 5) 
    ^  (A[15] >> 6) 
    ^  (A[15) >> 7)) & 1;

现在，您确实需要移位：

A[0] = (A[0] >> 1) | (A[1] << 7);
A[1] = (A[1] >> 1) | (A[2] << 7);
// and so on, until
A[14] = (A[14] >> 1) | (A[15] << 7);
A[15] = (A[15] >> 1) | (bit << 7);

您可以通过使用 uint32_t 或 uint64_t 而不是无符号字符来提高效率（取决于你的处理器字大小），但原理是一样的。

To calculate the bit to shift in you don't need to shift the whole array every time since you are only interested in one bit (note the & 1 at the end of the bit = line from Wikipedia).

The right shift amounts are:

128 - 128 =   0   => byte  0 bit 0
128 -  29 =  99   => byte 12 bit 3
128 -  27 = 101   => byte 12 bit 5
128 -   2 = 126   => byte 15 bit 6
128 -   1 = 127   => byte 15 bit 7

So,

bit = ((A[0] >> 0) 
    ^  (A[12] >> 3) 
    ^  (A[12] >> 5) 
    ^  (A[15] >> 6) 
    ^  (A[15) >> 7)) & 1;

Now, you really need to shift in the bit:

A[0] = (A[0] >> 1) | (A[1] << 7);
A[1] = (A[1] >> 1) | (A[2] << 7);
// and so on, until
A[14] = (A[14] >> 1) | (A[15] << 7);
A[15] = (A[15] >> 1) | (bit << 7);

You can make this a bit more efficient by using uint32_t or uint64_t instead of unsigned chars (depending on your processor word size), but the principle is the same.

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