What is 0xFF and why is it shifted 24 times?

Question

#define SwapByte4(ldata) \
   (((ldata & 0x000000FF) << 24) | \
   ((ldata & 0x0000FF00) << 8) | \
   ((ldata & 0x00FF0000) >> 8) | \
   ((ldata & 0xFF000000) >> 24))

What does that 0x000000FF represent? I know that decimal 15 is represented in hex as F, but why is it << 24?

Answer 1

You need to look at the 0x000000FF as a bitmask, i.e. where it's 1 the value of ldata will be taken and where it's 0 - 0 will be taken.

In order to understand the bitmask u need to convert it to binary, with hex it's very easy, every hex number is 4 binary digits, i.e.:

hex 0 = binary 0000 hex 1 = binary 0001 and so on.

Now to shifts: notice that the shift takes some data from the source, 8 bits exactly, and moves it to another location in the destination.

Now note that there's | i.e. OR operation on all the bitmask AND operations, i.e. zeroes will stay zeroes and in case there's '1' the result will contain one.

Hope it helps :)

Answer 2

Let's say data is a 32 bit number represented as 0x12345678 (each number is 4 bits in hex)

Data & 0x000000FF means keep only the last 8 bits (called a bit mask) = 0x00000078

The << 24 means move this value to the left 24 bits (78 starts at position 24 [0 index]) = 0x78000000

The | means logical or which in this case will just be an addition

Final result = 0x78563412

Read on logical manipulations

Answer 3

This kind of code tends to be used to swap things between big endian and little endian format. There is also a little trick that will convert a word in some known format (lets say, little endian) into whatever endianness the current machine happens to be, and vice versa. That would go something like this:

unsigned long littleEndian;
unsigned char* littleBytes = &littleEndian;
unsigned long result = 0;
for (i = 0; i < 4; i++)
    result += unsigned long(littleBytes[i]) << (8 * i);

This works (assuming I haven't messed it up) because regardless of how bytes are actually stored, shift left is guaranteed to shift towards more significant bits. Converting to a char* allows you to access the bytes in the order they are actually stored in memory. Using this trick you don't need to detect the machine endianness to read/write stuff in a known format. Admittedly you could also just use the standard functions (hton etc.) :P

(Note: You have to be a little careful and cast the char before shifting, otherwise it just overflows all over your shoes. Also, += isn't the only option, |= would probably make more sense but might be less clear if you aren't used to it, I'm not sure)