Question

In the following code, can the value of int be predicted ( how ? ), or it is just the garbage ?

union a
{
    int i;
    char ch[2];
};
a u;
u.ch[0] = 0;
u.ch[1] = 0;
cout<<u.i;
}

Answer 1

I would say that depends on the size of int and char. A union contains the memory of the largest variable. If int is 4 bytes and char[2] represents 2 bytes, the int consumes more memory than the char-array, so you are not initialising the full int-memory to 0 by setting all char-variables. It depends on your memory initialization mechanisms but basically the value of the int will appear to be random as the extra 2 bytes are filled with unspecified values.

Besides, filling one variable of a union and reading another is exactly what makes unions unsafe in my oppinion.

If you are sure that int is the largest datatype, you can initialize the whole union by writing

union a
{
    int i;
    char ch[2];
};

void foo()
{
    a u = { 0 };  // Initializes the first field in the union
    cout << u.i;
}

Therefore it may be a good idea to place the largest type at the beginning of the union. Althugh that doesn’t garantuee that all datatypes can be considered zero or empty when all bits are set to 0.