Expressing rotation where bits are represented in array

Question

I've got a series of bits, represented by pointers.

Bit* bits[32];

Now I need to implement the binary right rotate operator. I've got that the first part is a right shift of the same amount, which deals with all the bits which did not wrap around, but I'm not sure what to do about the ones which do wrap around.

How can I determine which bits end up in which positions in the end results?

So far, I've got

ExpressionHolder rotate(unsigned int other) const {
    ExpressionHolder out;
    out.e = new Expression;
    for(int i = 0; i < (32 - other); i++) {
        out.e->bits[i] = e->bits[i + other];
    }  
    for(int i = (32 - other); i < 32; i++) {
        out.e->bits[i] = e->bits[31 - i];
    }
    return out;
}

Is this correct?

You continue to surprise me with the occasional question that I'm sure you could have figured out yourself. Just feeling lazy today? — Ben Voigt
– Ben Voigt, Commented May 29, 2012 at 19:08
@BenVoigt: I distrust any intuition where the type system won't catch a mistake and I won't even find out if there is a mistake until I depend upon the results. I behave the same way with algorithms on float - no matter what shit you write, you always get a float back out and if the answer isn't what it should be, good luck working out why. — Puppy
– Puppy, Commented May 29, 2012 at 19:28

Praetorian · Accepted Answer · 2012-05-29 19:23:04Z

4

std::rotate_copy should do the trick

ExpressionHolder rotate( ExpressionHolder const& in, std::size_t count )
{
  ExpressionHolder out;
  out.e = new Expression();

  std::rotate_copy( in.e->bit, in.e->bit + 32 - count, in.e->bit + 32, out.e->bit );
  return out;
}

answered May 29, 2012 at 19:23

Praetorian

110k20 gold badges248 silver badges340 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Captain Giraffe · Accepted Answer · 2012-05-29 19:04:16Z

3

Quantum mechanics dictates that bits are always in a superposition of one and zero. So we can never be sure of the actual bit value of your pointer until we look at *e.bits[i].

But then after we look a straightforward

for(int i = 0; i < 32; i++)
     out.e->bits[i] = e->bits[(i + other) %32];

should do the trick.

answered May 29, 2012 at 19:04

Captain Giraffe

14.8k7 gold badges42 silver badges69 bronze badges

1 Comment

Puppy Over a year ago

Nice QM reference, but bits[i] only represents a bit. It is not an actual bit and does not have a defined value.

Collectives™ on Stack Overflow

Expressing rotation where bits are represented in array

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related