How to swap nibbles in C?

Question

How to swap the nibble bit positions of a number?

For example: 534, convert it into binary, the rightmost 4 bits has to be interchanged with the leftmost 4 bits and then make a new number with that.

Anyone know how to do this?

Solution 1

Start from the fact that hexadecimal 0xf covers exactly four bits. There are four nibbles in a 16-bit number. The masks for the nibbles are 0xf000, 0xf00, 0xf0, and 0xf. Then start masking, shifting and bitwise OR-ing.

Solution 2

Sean Anderson's bit twiddling guide has the following:

// swap nibbles ... 
v = ((v >> 4) & 0x0F0F0F0F) | ((v & 0x0F0F0F0F) << 4);

under the entry for Reverse an N-bit quantity in parallel in 5 * lg(N) operations.

Solution 3

1)

y = ((x >> 4) & 0x0f) | ((x << 4) & 0xf0);

2)

unsigned char swap_nibbles(unsigned char c)
{
     unsigned char temp1, temp2;

     temp1 = c & 0x0F;
     temp2 = c & 0xF0;
     temp1=temp1 << 4;
     temp2=temp2 >> 4;

     return(temp2|temp1); //adding the bits
}

3)

unsigned char nibbleSwap(unsigned char a)
{
    return (a<<4) | (a>>4);
}

Comments

[Admin]

Actually, what is the difference between 0x0F and 0xF0 ?

[Nikolai Fetissov]

0xf0 is 240 in decimal, while 0x0f is 15.

[Admin]

I saw this example from one site.. unsigned char c= 0x34; temp1 = c & 0x0F; temp2 = c & 0xF0; temp1=temp1 << 4; temp2=temp2 >> 4; return(temp2|temp1); What does that mean?

[Admin]

I saw this example from one site.. unsigned char c= 0x34; temp1 = c & 0x0F; temp2 = c & 0xF0; temp1=temp1 << 4; temp2=temp2 >> 4; return(temp2|temp1); What does that mean?

[JoeFish]

It means: assign the value 0x34 to c; store the bottom 4 bits of c in temp1; store the top 4 bits of c in temp2; shift the bits in temp1 4 places left; shift the bits in temp2 4 places right; OR temp2 and temp1 together to create an 8-bit value. Do some Google searching for bit manipulation and it will make more sense.

[sarnold]

Note that this is for reconstructing a 32 bit integer from an array of bytes. It makes an assumption about the endianness of the host processor and thus cannot be blindly used in programs -- there must be a corresponding decode_32le() routine that works on platforms of the opposite endianness in case this code is compiled on such a platform. It's also unrelated to nibble swapping, since it arranges bytes.

[subbupkb]

yes, your very right. hence forth i had mentioned its for 32 bit Endiness swap. thanks for more clarification.

[LtWorf]

You forgot to make sure the bitmask size is correct with sizeof, remembering that there are no size guarantees in C

[ramrunner]

would you mind fixing the code a bit with consistent indentation and having main actually return something? the code displayed in the site should be high quality and pass a code review cleanly if possible.