implicit assignment operators

Question

If I have an operator overload on my class, is the assignment version of the operator implicitly created as well?

class square{
   square operator+(const square& B);
   void operator=(const square& B);
};

As in, can I then call

square A, B;
A += B;

with the compiler implicitly deciding to call operator+ then operator=?

Answer 1

No, the operators aren't implicitly defined. However, boost/operators.hpp defines useful helper templates to avoid boiler-plate code. Example from their docs :

If, for example, you declare a class like this:

class MyInt
    : boost::operators<MyInt> {
    bool operator<(const MyInt& x) const;
    bool operator==(const MyInt& x) const;
    MyInt& operator+=(const MyInt& x);
    MyInt& operator-=(const MyInt& x);
    MyInt& operator*=(const MyInt& x);
    MyInt& operator/=(const MyInt& x);
    MyInt& operator%=(const MyInt& x);
    MyInt& operator|=(const MyInt& x);
    MyInt& operator&=(const MyInt& x);
    MyInt& operator^=(const MyInt& x);
    MyInt& operator++();
    MyInt& operator--(); };

then the operators<> template adds more than a dozen additional operators, such as operator>, <=, >=, and (binary) +. Two-argument forms of the templates are also provided to allow interaction with other types.

In addition, there is also support for implicitly "deducing" just a specific set of operators using arithmetic operator templates.