Question

Hello is have a question for a school assignment i need to :

Read a round number, and with the internal binaire code with bit 0 on the right and bit 7 on the left.

Now i need to change:
bit 0 with bit 7
bit 1 with bit 6
bit 2 with bit 5
bit 3 with bit 4

by example :

if i use hex F703 becomes F7C0
because 03 = 0000 0011 and C0 = 1100 0000
(only the right byte (8 bits) need to be switched.
The lession was about bitmanipulation but i can’t find a way to make it correct for al the 16 hexnumbers.

I`am puzzling for a wile now,

i am thinking for using a array for this problem or can someone say that i can be done with only bitwise ^,&,~,<<,>>, opertors ???

Answer 1

Study the following two functions:

bool GetBit(int value, int bit_position)
{
    return value & (1 << bit_position);
}

void SetBit(int& value, int bit_position, bool new_bit_value)
{
    if (new_bit_value)
        value |= (1 << bit_position);
    else
        value &= ~(1 << bit_position);
}

So now we can read and write arbitrary bits just like an array.

1 << N

gives you:

000...0001000...000

Where the 1 is in the Nth position.

So

1 << 0 == 0000...0000001
1 << 1 == 0000...0000010
1 << 2 == 0000...0000100
1 << 3 == 0000...0001000
...

and so on.

Now what happens if I BINARY AND one of the above numbers with some other number Y?

X = 1 << N
Z = X & Y

What is Z going to look like? Well every bit apart from the Nth is definately going to be 0 isnt it? because those bits are 0 in X.

What will the Nth bit of Z be? It depends on the value of the Nth bit of Y doesn’t it? So under what circumstances is Z zero? Precisely when the Nth bit of Y is 0. So by converting Z to a bool we can seperate out the value of the Nth bit of Y. Take another look at the GetBit function above, this is exactly what it is doing.

Now thats reading bits, how do we set a bit? Well if we want to set a bit on we can use BINARY OR with one of the (1 << N) numbers from above:

X = 1 << N
Z = Y | X

What is Z going to be here? Well every bit is going to be the same as Y except the Nth right? And the Nth bit is always going to be 1. So we have set the Nth bit on.

What about setting a bit to zero? What we want to do is take a number like 11111011111 where just the Nth bit is off and then use BINARY AND. To get such a number we just use BINARY NOT:

X = 1 << N   // 000010000
W = ~X       // 111101111
Z = W & Y

So all the bits in Z apart from the Nth will be copies of Y. The Nth will always be off. So we have effectively set the Nth bit to 0.

Using the above two techniques is how we have implemented SetBit.

So now we can read and write arbitrary bits. Now we can reverse the bits of the number just like it was an array:

int ReverseBits(int input)
{
    int output = 0;

    for (int i = 0; i < N; i++)
    {
        bool bit = GetBit(input, i); // read ith bit

        SetBit(output, N-i-1, bit); // write (N-i-1)th bit
    }

    return output;
}

Please make sure you understand all this. Once you have understood this all, please close the page and implement and test them without looking at it.

If you enjoyed this than try some of these:

http://graphics.stanford.edu/~seander/bithacks.html

And/or get this book:

http://www.amazon.com/exec/obidos/ASIN/0201914654/qid%3D1033395248/sr%3D11-1/ref%3Dsr_11_1/104-7035682-9311161