是否可以使用按位和受限操作符重写模(2 ^ n – 1)

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,则此操作将起作用.

编辑：该模式应该足够明显,看看如何将其推广到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次迭代应该足够了.

编辑2：

这是一个稍微更优化的版本,结合Paul R.的条件减法代码：

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