C++ add up all bytes in a structure

Question

Given a packed struct like this:

struct RSDPDescriptor {
    char Signature[8];
    uint8_t Checksum;
    char OEMID[6];
    uint8_t Revision;
    uint32_t RsdtAddress;
} __attribute__ ((packed));

How can I sum all of the individual bytes in it?

Answer 1

Here is some code that shows two ways to do it.

The first way is easier and more efficient, but will give the wrong result for a struct that doesn't have the packed attribute (since it will incorrectly include the padding bytes in its tally).

The second approach will work on any struct, padded or packed.

#include <stdio.h>
#include <stdlib.h>

template<typename T> int CountBytes(const T & t)
{
   int count = 0;
   const unsigned char * p = reinterpret_cast<const unsigned char *>(&t);
   for (int i=0; i<sizeof(t); i++) count += p[i];
   return count;
}

struct RSDPDescriptor {
    char Signature[8];
    unsigned char Checksum;
    char OEMID[6];
    unsigned char Revision;
    unsigned int RsdtAddress;
} __attribute__ ((packed));

int main(int, char **)
{
   struct RSDPDescriptor x;

   int byteCountFast = CountBytes(x);
   printf("Fast result (only works correctly if the struct is packed) is:  %i\n", byteCountFast);

   int byteCountSafe = CountBytes(x.Signature) + CountBytes(x.Checksum) + CountBytes(x.OEMID) + CountBytes(x.Revision) + CountBytes(x.RsdtAddress);
   printf("Safe result (will work even if there is padding) is:  %i\n", byteCountSafe);

   return 0;
}