Swap byte 2 and 4 in a 32 bit integer - c

Question

I had this interview question -
Swap byte 2 and byte 4 within an integer sequence.
Integer is a 4 byte wide i.e. 32 bits
My approach was to use char *pointer and a temp char to swap the bytes.
For clarity I have broken the steps otherwise an character array can be considered.
unsigned char *b2, *b4, tmpc;
int n = 0xABCD; ///expected output 0xADCB
b2 = &n; b2++;
b4 = &n; b4 +=3;
///swap the values;
tmpc = *b2;
*b2 = *b4;
*b4 = tmpc;
Any other methods?

Answer 1

int someInt = 0x12345678;
int byte2 = someInt & 0x00FF0000;
int byte4 = someInt & 0x000000FF;
int newInt = (someInt & 0xFF00FF00) | (byte2 >> 16) | (byte4 << 16);
To avoid any concerns about sign extension:
int someInt = 0x12345678;
int newInt = (someInt & 0xFF00FF00) | ((someInt >> 16) & 0x000000FF) | ((someInt << 16) & 0x00FF0000);
(Or, to really impress them, you could use the triple XOR technique.)
Just for fun (probably a tupo somewhere):
int newInt = someInt ^ ((someInt >> 16) & 0x000000FF);
newInt = newInt ^ ((newInt << 16) & 0x00FF0000);
newInt = newInt ^ ((newInt >> 16) & 0x000000FF);
(Actually, I just tested it and it works!)

Answer 2

You can mask out the bytes you want and shift them around. Something like this:
unsigned int swap(unsigned int n) {
unsigned int b2 = (0x0000FF00 & n);
unsigned int b4 = (0xFF000000 & n);
n ^= b2 | b4; // Clear the second and fourth bytes
n |= (b2 << 16) | (b4 >> 16); // Swap and write them.
return n;
}
This assumes that the "first" byte is the lowest order byte (even if in memory it may be stored big-endian).
Also it uses unsigned ints everywhere to avoid right shifting introducing extra 1s due to sign extension.

Answer 3

What about unions?
int main(void)
{
char tmp;
union {int n; char ary[4]; } un;
un.n = 0xABCDEF00;
tmp = un.ary[3];
un.ary[3] = un.ary[1];
un.ary[1] = tmp;
printf("0x%.2X\n", un.n);
}
in > 0xABCDEF00
out>0xEFCDAB00
Please don't forget to check endianess. this only work for little endian, but should not be hard to make it portable.