Converting a UINT32 value into a UINT8 array[4]

Question

My question is how do you convert a UINT32 value to a UINT8 array[4] (C/C++) preferably in a manner independent of endianness? Additionally, how would you reconstruct the UINT32 value from the UINT8 array[4], to get back to where you started?

Answer 1

You haven't really said what you mean by independent of endianness - it's unclear since the byte array must have some endianness. That said, one of the below must answer your requirements:

Given UINT32 v and UINT8 a[4]:

"Host" endian

(use the machine's native byte order):

UINT8 *vp = (UINT8 *)&v;
a[0] = vp[0];
a[1] = vp[1];
a[2] = vp[2];
a[3] = vp[3];

or:

memcpy(a, &v, sizeof(v));

or:

*(UINT32 *)a = v;

Big endian

(aka "network order"):

a[0] = v >> 24;
a[1] = v >> 16;
a[2] = v >>  8;
a[3] = v;

Little endian

a[0] = v;
a[1] = v >>  8;
a[2] = v >> 16;
a[3] = v >> 24;

Answer 2

E.g. like this:

UINT32 value;
UINT8 result[4];

result[0] = (value & 0x000000ff);
result[1] = (value & 0x0000ff00) >> 8;
result[2] = (value & 0x00ff0000) >> 16;
result[3] = (value & 0xff000000) >> 24;

Edit: added parenthesis (>> seems to have higher precedence than &)

Answer 3

If you don't want to code it yourself, you can use the C library function htonl() to convert the 32-bit int to network byte order. There is also the function ntohl() to convert them back to host order.

Once they're in network byte order, it's simply a matter of accessing the int/long as a byte array.

All in all that's are probably the most portable and tested way of achieving your goal.