The empty tuple/unit is not the same as void. It has exactly one possible value, void has zero possible values. A way to write it in Rust is `enum Void {}` (an enumeration with no options).

The point is that both void and Unit can only produce a side effect. In everything-is-an-expression based languages Unit is exactly equivalent to void in C*.

Unit is an empty tuple:

https://en.wikipedia.org/wiki/Tuple

It's the same thing. It's a type with only a single value.

`void` in C does not have a value. You can't make a variable and put a void in it because there is no such object as "void".

That is an artificial restriction if C. Also see how ! In Rust is also loosing it's artificial restrictions.

Indeed, and there has been some talk of removing this restriction in C++ because it makes certain kinds of metaprogramming a lot more cumbersome than they should be.

Yes, and () isn't void; the poster above me is wrong. Void is...the absence of a type? It's awkward.

However, () and Unit are.

It's not the same thing, a value of type Unit can only produce a side effect (or do nothing at all).

How is that different from a value of type "empty tuple"?

True, neither are useful as values, though Unit typically conveys programmer intent (to produce a side effect), whereas the empty tuple is, in Scala at any rate, quite rare.

The empty tuple:

    scala> val empty = Tuple1(())
    empty: (Unit,) = ((),)

vs. Unit:

    scala> val empty = ()
    empty: Unit = ()

Isn't that a tuple containing a unit, and therefore not empty?

It's the closest you can get in Scala to represent an empty value whose type is `Tuple`.

The difference between an empty tuple (also known as unit) and void becomes obvious when you deal with vaguely complex trait impls. For example, if you have a trait:

    trait Foo {
        type ErrorType;
        fn bar() -> Result<u8, Foo::ErrorType>;
    }

How would you specify that your type implements Foo in such a way that bar() cannot return an error? If you were to implement it using the empty tuple (unit), like this, it could actually return an error:

    struct Abc{}
    impl Foo for Abc {
        type ErrorType = ();
        fn bar() -> Result<u8, ()> {
            Err(()) // oops, we don't want to be able to do that!
        }
    }

Instead, you can use Void here:

    enum Void{}
    struct Xyz{}
    impl Foo for Xyz {
        type ErrorType = Void;
        fn bar() -> Result<u8, Void> {
            // No way to create a Void, so the only thing we can return is an Ok
            Ok(1)
        }
    }

Aside from the obvious power to express intent (can we return an error without any information attached, or can we not error at all?), this would allow an optimizing compiler to assume that the result of Xyz::bar() is always a u8, allowing it to strip off the overhead of the Result:

    fn baz<F: Foo>(f: F) {
        match f.bar() {
            Ok(v) => println!("{}", v),
            Err(e) => panic!("{}", e)
        };
    }
    ...
    baz(Xyz); // The compiler can notice that f.bar() can never return a Result::Err, so strip off the match and assume it's a Result::Ok

A super-smart compiler would even make sure it's not storing the data for "is this an Ok or Err" in the Result<u8, Void> at all.

Finally, similarly, you can specify that certain functions are uncallable by having them take Void as a parameter.

I don't think anyone is claiming Void in Scala/Haskell/Rust is equivalent to Unit in Scala/Haskell/Rust.

The question here was Unit vs empty tuple.

Up-thread was the question of whether C "void" is more like S/H/R Unit or S/H/R Void.

Upthread was:

> If Rust follows Scala then `()` is not the empty tuple, but rather Unit (void in C*).

The implication is that () == Unit == void. The empty tuple and unit are essentially equivalent aside from name, void is something else.

Regardless of who is right, you are arguing the wrong bit of it. I contend that, to a person, everyone saying "C void is Unit" doesn't think C void is Void. Arguing that Void is not Unit is just completely spurious. Of course Void is not Unit. No one disagrees.

In truth, C void is not exactly either Void or Unit. Like Void, you can't exactly make one... but you can call functions declared to take it and write functions that return it, and really it just means "I have no information to pass" - which is more like Unit.

Type names are irrelevant here. Unit could be called "Tuple0", or be defined as a synonym of a type named such. The semantics are identical.

> For historical reasons and convenience, the tuple type with no elements (()) is often called ‘unit’ or ‘the unit type’.

https://doc.rust-lang.org/reference.html

void in C can't be created, used, or passed around - () can.

So in Rust, the equivalent of void* with references instead looks like:

    enum Void{}
    fn foo(v: &Void){ ... }
    let bar : Void;
    fn abc() -> &Void{ ...}
    fn xyz() -> Void{ ... }

You can't create a Void, and you can't cast to it either - it's not a bottom type. So you can't put anything in bar. And as a result, you can't create a reference to a Void, so you can't call foo. And abc and xyz just can't be implemented in the first place.

On the other hand, you can do all of these just fine:

    fn foo(v: ()){ ... }
    fn bar(v: &()){ ... }
    ...
    let v = ();
    bar(&v);
    foo(v);

The fact that you can create and use an empty tuple as a value shows that it is not equivalent to Void.

(All statements here are made within the safe subset of the language - unsafe allows access to intrinsics that would allow a Void to be made, and a reference to Void.)

void * isn't related to void (or this discussion), they just reused the keyword.