How operator overloading in c++ works -

i going through of theory of assignment operator of c++.

let's say

class myclass { private:    t1 member1;    t2 member2; public:    // default copy assignment operator assigns object via memberwise copy    myclass & operator=(const myclass & rhs) {       member1 = rhs.member1;       member2 = rhs.member2;       return *this;    } ...... } 

and

c7 = c6; //member wise copy assignment. 

here returning reference object during assignment operation , assigning new object c7.

however if code this:

    int a=12;     int &b=a;     int c=&b;        //error::invalid conversion ‘int*’ ‘int’ 

why different above case??

the declaration

int& b = a; 

declares b reference, , makes reference variable a.

the declaration

int c = &b; 

declares c normal int variable, , try initialize pointer int. ampersand & different things depending on context.

---

as copy-assignment operator, it's not called when initialize variable in declaration, instead it's copy-constructor being called.

---

when have copy-assignment operator, , e.g.

myclass a, b;  = b;  // copy-assignment operator called 

the compiler replaces assignment following call

a.operator=(b); 

in other words, it's normal member-function call, other.