Swap the first 8 bits and the last 8 bits of 16-bit value

Question

For example, 00000001 11001100 becomes 11001100 00000001;

unsigned short swap_bytes(unsigned short x) {

}

Answer 1

Method 1:

extract the high byte hibyte = (x & 0xff00) >> 8;
extract the low byte lobyte = (x & 0xff);
combine them in the reverse order x = lobyte << 8 | hibyte;

Method 2:

typedef union mini
{
    unsigned char b[2];
    unsigned short s;
} micro;

unsigned short swap_bytes(unsigned short x) 
{
    micro y;
    y.s = b;
    unsigned char tmp = y.b[0];
    y.b[0] = y.b[1];
    y.b[1] = tmp;
    return y.s
}

Comments

Can  u explain your code for beginner please...

Answer 2

Mask first 8 bits so you will get second 8 bits and mask second 8 bits so you will get first 8 bits. Now do OR for these two then you will get swapped number.

(num & 0x00FF ) || (num & 0xFF00 ) .

Comments

I see. Well I think U mean "return (x & 0x00FF ) || (x & 0xFF00 );" Right?

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yes............