Question

class Currency
{
public:
    explicit Currency(unsigned int value);
    // method form of operator+=
    Currency &operator +=(const Currency &other); // understood!
    ...
};

The following code shows an equivalent API using a free function version of the operator:

class Currency
{
public:
    explicit Currency(unsigned int value);
    ...
};

// free function form of operator+=
Currency &operator +=(Currency &lhs, const Currency &rhs); // ???

Question1> Why should the free function return Currency& instead of Currency?
Is this a good practice?

Question2> In the implementation, which variable should be used to return, lhs or rhs?

Answer 1

The standard behavior of operator+= is to increment the lhs by the rhs and return a reference to the lhs.

In the member function, the lhs is the calling object, and accordingly, it should return a reference to itself. You seem to expect the free function to behave differently than the member function. Why?