How do I simulate a bitwise rotation of a 64-bit (unsigned) integer in JavaScript?

Question

I need to perform a circular left shift of a 64-bit integer in JavaScript. However:

• JavaScript numbers are doubles
• JavaScript converts them to 32-bit signed ints when you start with the << and the >> and the >>> and the ~ and all of the bit-twiddling business. And then it's back to doubles when you're done. I think.
• I don't want the sign. And I definitely don't want the decimal bits. But I definitely do want 64 bits.

So, how do I perform a bitwise left rotation of a 64-bit value?

Answer 1

I believe so, though not the most efficient way, convert the number to a string in binary form (64-bits), use substring to move the char at the beginning and append it to the end (for left rotation) and convert the binary form back to number. I am sure you can figure out how to convert a decimal number to its binary form into a string and back.

Answer 2

The only way I think it can be done is to create an int64 class which internally contains two 32 bit integers and performs shifting by carrying between them.

Answer 3

Keep your 64-bit number as separate high and low halves. To rotate left N when N < 32:

hi_rot = ((hi << N) | (lo >>> (32-N))) & (0xFFFFFFFF)

lo_rot = ((lo << N) | (hi >>> (32-N))) & (0xFFFFFFFF)

If N >= 32, then subtract 32 from N, swap hi and lo, and then do the above.

Answer 4

Here's a values based rotate.

double d = 12345678901.0;
// get high int bits in hi, and the low in
int hi = (int)(d / 16.0 / 16.0 / 16.0 / 16.0);
int low = (int)d;

int rot = 3; // thus * 8
int newhi = (low >> (32 - rot)) | (hi << rot);
int newlow = (hi >> (32 - rot)) | (low << rot);

double newdouble = ((double)hi * 16.0 * 16.0 * 16.0 * 16.0) + (double)low;