C++ assignment operator and overloading - c++

Question

I was wondering why the following code is being compiled successfully.
#include <iostream>
#include <cassert>
using namespace std;
class Integer{
private:
int i;
public:
Integer(int value):i(value){}
// unary or binary
const Integer operator+(const Integer& rv){
cout << "operator+"<< endl;
return Integer(i + rv.i);
}
// Binary Operator
Integer& operator+=(const Integer& rv){
cout << "operator+=" << endl;
i += rv.i;
return *this;
}
ostream& operator<<(ostream& lv){
lv << i;
return lv;
}
friend ostream& operator<< (ostream& lv, Integer& rv);
};
ostream& operator<< (ostream& lv,Integer& rv){
return lv << rv.i;
}
int main(){
cout << "using operator overloading"<< endl;
Integer c(0), a(4), b(5);
Integer d = 8;
c = a + b;
cout << c << endl;
cout << d << endl;
}
I dont understand why d = 8 is possible. d is a user defined type. I did not overloaded the assignment oeprator for the Integer class. Is there a default overloaded operator?

Answer 1

You have not declared the Integer constructor explicit, so it acts as an implicit conversion from int to Integer.
If you declare your constructor
explicit Integer(int value);
the compiler fires the error:
error: conversion from ‘int’ to non-scalar type ‘Integer’ requested

Answer 2

Your constructor defines conversion from int to Integer
Integer(int value):i(value){}

Answer 3

I dont understand why d = 8 is possible. d is a user defined type.
Just instrument your code a little more and also trace the ctor calls.
#include <iostream>
#include <cassert>
using namespace std;
class Integer{
private:
int i;
public:
Integer(int value):i(value){
cout << "Integer(int) called with " << value << endl;
}
// unary or binary
const Integer operator+(const Integer& rv){
cout << "operator+"<< endl;
return Integer(i + rv.i);
}
// Binary Operator
Integer& operator+=(const Integer& rv){
cout << "operator+=" << endl;
i += rv.i;
return *this;
}
ostream& operator<<(ostream& lv){
lv << i;
return lv;
}
friend ostream& operator<< (ostream& lv, Integer& rv);
};
ostream& operator<< (ostream& lv,Integer& rv){
return lv << rv.i;
}
int main(){
Integer d = 8;
cout << d << endl;
}
output:
Integer(int) called with 8
8
live at Coliru's
The presumed assignment of the immediate 8 in fact causes the non-explicit Integer(int) ctor to be called with 8 as its argument.