c - 返回 x，其中从位置 p 开始的 n 位设置为 y 的最右边 n 位，其他位保持不变

Question

我的解决方案

get the rightmost n bits of y
a = ~(~0 << n) & y

clean the n bits of x beginning from p
c = ( ~0 << p | ~(~0 << (p-n+1))) & x

set the cleaned n bits to the n rightmost bits of y
c | (a << (p-n+1))

这是相当长的陈述。我们有更好的吗？

 i.e
x = 0 1 1 1 0 1 1 0 1 1 1 0
p = 4
y = 0 1 0 1 1 0 1 0 1 0 
n = 3

the 3 rightmost bits of y is 0 1 0 
it will replace x from bits 4 to bits 2 which is 1 1 1

Answer 1

我写过类似的:

 unsigned setbits (unsigned x, int p, int n, unsigned y)
{
  return (x & ~(~(~0<<n)<<(p+1-n)))|((y & ~(~0<<n))<<(p+1-n));
}