Make a struct have non-padded size

Question

So, I want to be able to write binary data in C that alternates between a single byte and a 32-bit signed integer, and have a struct type that can be used for this with pointer incrementing.

So far, this is my struct

struct MarkedInt
{
  unsigned char mark;
  int value;
};

And this is my code

void clean_obj(struct MarkedInt* mi)
{
  mi->mark = 0;
  mi->value = 0;
}

int main(void) {

  struct MarkedInt a;
  clean_obj(&a);
  a.mark = 5;
  a.value = 500;

  unsigned char* printer = (unsigned char*)(&a);
  printf("sizeof a is %lu\n", sizeof(a));
  for(int i = 0; i< sizeof(a);i++)
  {
    printf("%u\n", printer[i]);
  }
  return 0;
}

However, the result of the program is this

sizeof a is 8
5
4
64
0
244
1
0
0

So it sets the size of my struct MarkedInt at 8, making the mark member 4 bytes instead of 1. This causes some unwanted padding, where a pointer to the struct used to read binary data would be reading 8 bytes at a time and not 5.

My desired outcome is this

unsigned char buf[] = {/*mark*/ 5, /*int value*/ 1, 0, 0, 0};
struct MarkedInt* reader = buf;
printf(" mark: %u, value: %d\n", reader->mark, reader->value);

Is it possible to do this in c? Or is simply reading the byte marker and int separately recommended?