是否可以使用按位运算符和受限运算符重写模数(2 ^ n - 1)

Question

对于unsigned int x,是否可以仅使用以下运算符(加上没有循环,分支或函数调用)来计算x%255(或者通常为2 ^ n - 1)？

!,~,&,^,|,+,<<,>>.

Answer 1

是的,这是可能的.对于255,可以按如下方式完成:

unsigned int x = 4023156861;

x = (x & 255) + (x >> 8);
x = (x & 255) + (x >> 8);
x = (x & 255) + (x >> 8);
x = (x & 255) + (x >> 8);

//  At this point, x will be in the range: 0 <= x < 256.
//  If the answer 0, x could potentially be 255 which is not fully reduced.

//  Here's an ugly way of implementing: if (x == 255) x -= 255;
//  (See comments for a simpler version by Paul R.)
unsigned int t = (x + 1) >> 8;
t = !t + 0xffffffff;
t &= 255;
x += ~t + 1;

// x = 186

如果unsigned int是32位整数,这将起作用.

编辑:模式应该足够明显,以了解如何将其推广到2^n - 1.您只需要弄清楚需要多少次迭代.对于n = 8和32位整数,4次迭代就足够了.

编辑2:

这是一个稍微更优化的版本与Paul R.的条件减法代码相结合:

unsigned int x = 4023156861;

x = (x & 65535) + (x >> 16);     //  Reduce to 17 bits
x = (x & 255) + (x >> 8);        //  Reduce to 9 bits
x = (x & 255) + (x >> 8);        //  Reduce to 8 bits
x = (x + ((x + 1) >> 8)) & 255;  //  Reduce to < 255