Store a signed long int (32bit) as 4 octets?

Question

I managed to get a unsigned long int octets-representation (BE) by reading IPv4 methods, and I managed to read about how signed integers are using the MSB as the sign indicator, which makes 00 00 00 00 to be 0, while 7F FF FF FF is 2147483647.

But I can't manage how to do the same for signed long integers?

#include <stdio.h>
#include <string.h>

int main (void)
{
    unsigned long int intu32;
    unsigned char octets[4];

    intu32 = 255;

    octets[3] = (intu32) & 255;
    octets[2] = (intu32 >> 8) & 255;
    octets[1] = (intu32 >> 16) & 255;
    octets[0] = (intu32 >> 24) & 255;
    printf("(%d)(%d)(%d)(%d)\n", octets[0], octets[1], octets[2], octets[3]);

    intu32 = (octets[0] << 24) | (octets[1] << 16) | (octets[2] << 8) | octets[3];
    printf("intu32:%lu\n", intu32);

    return 0;
}

Thanks in advance, Doori bar

Answer 1

There is no difference. You can always serialize/deserialize signed integers as if they are unsigned, the difference is only in the interpretation of the bits, not in the bits themselves.

Of course, this only holds true if you know that the unsigned and signed integers are of the same size, so that no bits get lost.

Also, you need to be careful (as you are) that no intermediary stage does any unplanned sign-extension or the like, the use of unsigned char for individual bytes is a good idea.

Answer 2

You are probably confused that it is common practise (and applied in ix86 processors) to encode negative values using twos complement encoding. This means that the hex notation of 4 byte -1 is 0xffffffff. The reason this encoding is used is that by taking into account automatic overflow adding 2 0x00000002 to -1 will yield the correct result (0x00000001).

Answer 3

Do you want something like this? It would be helpul (as Vicki asked) if you could provide what you have and what you want to get.

#include <stdio.h>
#include <string.h>

int main (void)
{
    union{
        long int intu32;
        char octets[4];
    } u;
    u.intu32 = 255;

    printf("(%d)(%d)(%d)(%d)\n", (int) u.octets[3], (int) u.octets[2], (int) u.octets[1], (int) u.octets[0]);

    printf("intu32:%lu\n", u.intu32);

    return 0;
}