Do Assignment Operator Conversion

    You will find interesting. I want to convert from object B to object
A and I have to use vector.

vector< A > a;
vector< B > b;
a = b;

    You can't do that because vector does not have assignment operator
implementation. What will you do?
    Create class A and class B. Both class A and class B are derived
from vector. All the functions of vector are inherited into class A
and class B.

template< typename U >
class B;

template< typename T >
class A : public vector< T > {
public:
    template< typename U >
    A< T >& operator =( const B< U >& );
};

template< typename T >
class B : public vector< T > {
public:
};

template< typename T >
template< typename U >
A< T > &A< T >::operator =( const B< U > &right ) {
    A< T >::iterator iterA = begin();
    B< U >::const_iterator iterB = right.begin();

    while( iterA != end() ) {
        *iterA = *iterB; // sample only
// convertData( iterA, iterB ); // not implemented yet
        iterA++;
        iterB++;
    }

    return *this;
}

int main() {
    A< int > a;
    B< int > b;

    a.push_back( 1 );
    a.push_back( 2 );
    a.push_back( 3 );
    a.push_back( 4 );

    b.push_back( 10 );
    b.push_back( 20 );
    b.push_back( 30 );
    b.push_back( 40 );

    a = b;

    return 0;
}

    You already know that string is an example. It has a function to
convert from char* to string.
    What if you want to convert from string to any type of object?
    Think of class A is to be named pixelString and class B is to be
named string. You can do either two ways.

pixelString pixelStrName_A = =93There is a cat sitting on the fence.=94;

or=85

pixelString pixelStrName_A;
string strName_B = =93There is a cat sitting on the fence.=94;
pixelStringName_A = strName_B;

    Each character in the string will be converted to pixelString before
pixelString calls a function to draw 16 x 16 bitmap. You can modify
string and then bitmap will be redrawn automatically.