Understanding Character array initialization and data storing

I am trying to understand how character arrays are initialized and how are the data stored.

I did a test and this is how it goes

struct test
{
    char one  [2];
    char two  [2];
    char three[2];
};

test * testptr;

char testchar[] = "hello";

testptr = reinterpret_cast<test*>(testchar);

cout << testptr->one << endl;
cout << testptr->two << endl;
cout << testptr->three << endl;

I was expecting to see something like:

he
ll
o

However, when i complile, I get:

hello
llo
o

I do not understand how is one and twocontains more character than the size itself, when i do a sizeof for the variables, they all turn out to be 2, just as the size it was initialized.

And if a cout like this is done:

for (int i = 0; i < 6; i++)
{
    cout << testptr->one[i] << endl;
}

The result is:

h
e
l
l
o

If i go more than 6, it would be either empty or rubbish.

I know i can use memcpy to allocate nicely the amount bytes into the array, but i'm trying to understand how does a fixed size character array, able to store more that what it can hold.

Solution

When you reinterpret_cast the char[] to struct*, the pointers to one, two and three point to the 0th, 2nd, and 4th character respectively in the input string (with a gap of two since the size of each variable is two, so something like ptr, ptr + 2, ptr + 2 + 2).

h e l l o
^   ^   ^
|   |   |
0   2   4

Also, a char* in C++ is terminated by a NULL character ('\0'). Hence, when you print the value of one, two and three, we see all the characters from the beginning until the end is encountered.