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How to write this C expression in J?

How to write this C expression in J ? (where x is an integer and a is a temporary variable)

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

Strike>.

Edit:

In a more readable form:

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

Without testing, the basic transcription would be something like this:

 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. )

But there may be a reasonable approach.

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Here is a small analysis (versions of the "readable form"):

 usnigned int nx = ~x; // I suppose it 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

It seems that the algorithm is looking for the first 2 (starting with LSB) consecutive bits 0 in the variable x . If they are not, then the result is 0. If they are found, say, in position PZ , then the result will contain two bits: PZ and PZ+1 .

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