c++ - Usage of `void()` in a comma-separated list?

Question

I was looking into a piece of code written by others when I saw this:

a(), void(), b();

where both a and b are instances of a user-defined template class, which is intended to act like a function by overloading operator() that returns the calling instance itself.

Part of the class:

template <typename T>
class SomeClass{
    public:
    SomeClass& operator()(void);
    const SomeClass& operator()(void) const;
}

The return statements for both overloads are the following:

template <typename T>
SomeClass<T>& SomeClass<T>::operator()(void){
    // do stuff
    return *this;
}

template <typename T>
const SomeClass<T>& SomeClass<T>::operator()(void) const{
    // do stuff
    return *this;
}

What does the void() between them do? I feel it strange.

question from:https://stackoverflow.com/questions/46198648/usage-of-void-in-a-comma-separated-list

Answer 1

The void() prevents an overloaded operator, from being called (where one of the parameters is of the type SomeClass<T>), as such an overload can't have a parameter of type void.

You will most often see this used in templates, and is used in variadic pack expansions:

// C++11/14:
int unpack[] = {0, (do_something(pack), void(), 0)...};
// C++17 (fold expression):
(void(do_something(pack)), ...);

Where an overloaded operator, could ruin the sequencing guarantees of the language.