How do I write this C expression in J?

Answer 1

Here is a small analyzis (of the "readable form" version):

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

It seems that the algorithm is looking for the first 2 (starting from LSB) consecutive 0 bits in the x variable. If there aren't any, then the result is 0. If they are found, say on position PZ, then the result will contain two set bits: PZ and PZ+1.

Answer 2

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 could be a smarter approach.