Construction of a void Type?

Question

I was given a piece of code that uses void() as an argument. The code doesn\'t compile... obviously?

Can we instantiate anything of type void

Answer 1

The expression void() is a prvalue of type void and can be used anywhere such an expression may be used, which [basic.fundamental]/9 helpfully provides a list:

• As an expression-statement: void();
• As the second or third operand of a conditional operator: true ? throw 1 : void()
• As an operand of the comma operator: ++it1, void(), ++it2
• As the operand of decltype or noexcept: using my_void = decltype(void()); static_assert(noexcept(void()), "WAT");
• In a return statement of a function returning (possibly cv-qualified) void: const void f() { return void(); }
• As an operand of an explicit conversion to (possibly cv-qualified) void: static_cast(void())

An expression of type void can also be used as the operand of typeid, but void() in particular would be parsed as a type, not an expression, in this context.