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如何用J写这个C表达式?

j c

Margus · 技术社区 · 15 年前

我该怎么写这个C表达式呢 J ? (其中 x a 是临时变量)

((a= ~x & (~x >> 1)) ^= a ? 0 : (a ^ (a & (a - 1))) | (a ^ (a & (a - 1))) << 1);

.

编辑:

以更易读的形式:

int a = (~x) & ((~x) >> 1); if (a == 0) return 0; int b = a ^ (a & (a - 1)); return b | (b << 1);

2 回复 | 直到 15 年前

MPelletier 15 年前

如果没有测试,基本的转录会是这样的:

Shift =: (33 b.)
And   =: (17 b.)
Not   =: (26 b.)
Xor   =: (22 b.)
Or    =: (23 b.)

BastardFunction =: 3 : 0
  a =. (Not y) And (_1 Shift (Not y))
  if. a do.
    b =. a  Xor (a And a - 1)
    (1 Shift b) Or b
  else.
    0
  end.
)

ruslik 15 年前

下面是一个小分析(即“可读形式”版本):

usnigned int nx = ~x;   // I suppose it's unsigned 
int a = nx & (nx >> 1); 
// a will be 0 if there are no 2 consecutive "1" bits.
// or it will contain "1" in position N1 if nx had "1" in positions N1 and N1 + 1
if (a == 0) return 0;   // we don't have set bits for the following algorithm
int b = a ^ (a & (a - 1));  
// a - 1 : will reset the least 1 bit and will set all zero bits (say, NZ) that were on smaller positions
// a & (a - 1) : will leave zeroes in all (NZ + 1) LSB bits (because they're only bits that has changed
// a ^ (a & (a - 1)) : will cancel the high part, leaving only the smallest bit that was set in a
// so, for a = 0b0100100 we'll obtain a power of two: b = 0000100
return b | (b << 1);    
// knowing that b is a power of 2, the result is b + b*2 => b*3

似乎算法在寻找前2个(从LSB开始)连续的 0 x 如果没有,那么结果是0。如果他们被发现了,说在位置上 PZ ,则结果将包含两个设置位: PZ公司 和 PZ+1 .