如何在J中编写此C表达式? (其中x
是输入整数,a
是临时变量)
<击> 撞击>
<击>((a= ~x & (~x >> 1)) ^= a ? 0 : (a ^ (a & (a - 1))) | (a ^ (a & (a - 1))) << 1);
击> <击> 撞击>
以更易阅读的形式:
int a = (~x) & ((~x) >> 1);
if (a == 0) return 0;
int b = a ^ (a & (a - 1));
return b | (b << 1);
答案 0 :(得分:5)
没有测试,基本的转录将是这样的:
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.
)
但可能会采用更聪明的方法。
答案 1 :(得分:3)
这是一个小分析(“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
似乎该算法正在查找0
变量中的前2个(从LSB开始)连续x
位。
如果没有,则结果为0。
如果找到它们,请说明位置PZ
,那么结果将包含两个设置位:PZ
和PZ+1
。