Re: Overloading Operator[] through 'this' pointer?
Immortal Nephi <Immortal_Nephi@hotmail.com> writes:
Can overloading operator[] be connected to any member functions
through ??this?? pointer? I do not want to insert assignment between
array object and integer value. I prefer to place integer value in
the member function??s parameter.
For example:
class Array
{
public:
Array( int size ) : pData( new unsigned char[ size ] )
{}
~Array()
{
delete [] pData;
}
int &operator[]( int index )
{
return pData[ index ];
}
Array &Set_LowNibble( unsigned char val, int index )
{
pData[ index ] &= 0xF0;
pData[ index ] |= (val & 0x0F );
return *this;
}
Array &Set_HighNibble( unsigned char val, int index )
{
pData[ index ] &= 0x0F;
unsigned char temp = val;
temp &= 0x0F;
temp <<= 4;
pData[ index ] |= temp;
return *this;
}
private:
unsigned char *pData;
};
int main()
{
Array array( 4 );
array[ 0 ] = 0x1E; // Always valid code
// Not valid code but you find another way
array[ 1 ].Set_LowNibble( 0x0D, 1 ).Set_HighNibble( 0x02, 1 );
return 0;
}
The following is just an idea to think about. I have not given very
much consideration to the details and I am not completely sure why you
are wanting it to work in this way.
class Byte
{
public:
Byte(unsigned char& c) : c_(c) { }
Byte& operator=(unsigned char val)
{
c_ = val;
return *this;
}
Byte& Set_LowNibble(unsigned char val)
{
c_ &= 0xF0;
c_ |= (val & 0x0F);
return *this;
}
Byte& Set_HighNibble(unsigned char val)
{
c_ &= 0x0F;
unsigned char temp = val;
temp &= 0x0F;
temp <<= 4;
c_ |= temp;
return *this;
}
operator unsigned char&() { return c_; }
private:
unsigned char& c_;
};
class Array
{
public:
Array(int size)
: pData(new unsigned char[size])
{ }
~Array()
{
delete [] pData;
}
Byte operator[](int index)
{
return pData[index];
}
private:
unsigned char *pData;
};
int main()
{
Array array(4);
array[0] = 0x3C;
array[1].Set_LowNibble(0x0D).Set_HighNibble(0x02);
std::cout << array[0] << '\n';
std::cout << array[1] << '\n';
unsigned char& c_ref = array[1];
c_ref = array[0];
std::cout << array[0] << '\n';
std::cout << array[1] << '\n';
return 0;
}
/**
* Output:
* <
* -
* <
* <
*/
Regards
Paul Bibbings