How to define one side operators in c++?

Question

How to define one side operators like ++ or -- in c++?

for example we want to define a## to do (a % 45) + 2 [It is just an example]

There is no `##` operator in C++. You can't make up custom operators, only [overload predefined operators](https://en.cppreference.com/w/cpp/language/operators). As for `++` and `--`, they are formally known as [increment/decrement operators](https://en.cppreference.com/w/cpp/language/operator_incdec) — Remy Lebeau, Jun 28 '19 at 08:00
You can't *add* any operators at all, and you can't overload operators for primitive types. — molbdnilo, Jun 28 '19 at 08:22
To define `a##`, you'll need to create your own language. Your new language could be C++ like. Creating a language is a big undertaking. Toy languages can take a few months to write; (empirically) industrial strength languages take a decade or more. — Eljay, Jun 28 '19 at 11:24
The common name for the "one-sided" operators would be **_unary_** operators — MSalters, Jun 28 '19 at 14:05

score 3 · Accepted Answer · answered Jun 28 '19 at 07:59

There is operator ++() (prefix increment operator)
and operator ++(int) (postfix increment operator)

Same for operator --.

class Example
{
public:
   int a = 0;

   Example& operator++() { a = (a % 45) + 2; return *this; } // ++ex;
   Exampleoperator++(int) { Example tmp = *this; ++(*this); return tmp; } // ex++;
};

There is no operator ##.

How to define one side operators in c++?

1 Answers1