Help overloading assignment operator in C++?

Question

Trying to overload the assignment operator in C++

Point& Point::operator=(const Point& p)
{
if(this != &p)
{
delete [] label; //Deletes label
label = new char(strlen (*(p.label)));
strcpy(label, (*(p.label)));
x = (*(p.x)); //Copies x.
y = (*(p.y)); //Copies y
}
return *this; //Returns *this.
}

h:\point\point\point\point.cpp(64) : error C2664: 'strlen' : cannot convert parameter 1 from 'char' to 'const char *'
Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
h:\point\point\point\point.cpp(64) : fatal error C1903: unable to recover from previous error(s); stopping compilation

With line 64 commented out:

h:\point\point\point\point.cpp(65) : error C2664: 'strcpy' : cannot convert parameter 2 from 'char' to 'const char *'
Conversion from integral type to pointer type requires reinterpret_cast, C-style cast or function-style cast
h:\point\point\point\point.cpp(66) : error C2440: '=' : cannot convert from 'double' to 'double *'
h:\point\point\point\point.cpp(67) : error C2440: '=' : cannot convert from 'double' to 'double *'

Answer 1

label = new char(strlen (*(p.label)));

strcpy(label, (*(p.label)));

x = (*(p.x)); //Copies x.

y = (*(p.y)); //Copies y



The code above replace with



label = new char(strlen ((p.label)));

strcpy(label, ((p.label)));

x = p.x; //Copies x.

y = p.y; //Copies y

Answer 2

Ok, The line

Point& Point::operator = (const Point& p)

This means that your argument to the function is passed by reference, and returns a reference.

A reference argument is used like a regular argument, no special treatment required.

I believe the assignment operator is redefined as

Point& Point::operator = (const Point& p)

The complete function would be

Point& Point::operator = (const Point& p)

{

if (this == &p)

return *this;

char *temp;

double dx, dy;

temp = new char[strlen(p.label) + 1];

strcpy(temp, p.label);

dx = p.x;

dy = p.y;

x = dx;

y = dy;

delete label;

label = new char[strlen(temp) + 1];

strcpy(label, temp);

delete temp;

return *this;

}

Answer 3

you may declare the class yet with diverse information kinds. this is how the compiler is familiar with what functionality to apply. That way once you declare the class of a particular type, it purely makes use of the class matching the information type. consequently, overloaded.

Answer 4

????