Are there any null bytes after the `char` in the structure?

I'm trying to make sure I understand what the hidden assumptions are here.

This code gives correct results.

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

struct branch
{
    char flag;      //value
    struct branch *l; //left child
    struct branch *r; //right child
};

struct branch c={'c',NULL,NULL};
struct branch e={'e',NULL,NULL};
struct branch f={'f',NULL,NULL};
struct branch b={'b',&c,NULL};
struct branch d={'d',&e,&f};
struct branch a={'a',&b,&d};

void preorder(struct branch* t)
{
    printf(&t->flag);    //Seems ugly and errorprone
    if(t->l) preorder(t->l);
    if(t->r) preorder(t->r);
}

int main()
{
    preorder(&a);
}

      

        
        
        
      

    

The output, as expected, is abcdef

Can someone confirm my suspicions that this only works because two things:

• structure members are aligned on n-byte boundaries (n! = 1) (n = 4 seems to be when requesting sizeof () - s)
• bytes not used by the first member (which is char
  
  ) before the n-byte boundary are zeroed out.

I see no other explanation for printf to work correctly otherwise, since it expects a null terminated char [].

Also, is it reasonable to do things like this (outside of a unit-code situation where optimizations might outweigh readability and portability concerns), i.e. Are these assumptions more or less universal?

First week of intermittent messing with C, so I'm pretty green.