In C++17 can an if statement with an initializer be used to unpack an optional?

There is not, and cannot possibly be, such an Unpack() function.

But you could certainly do:

if (std::optional<int> o = GetOptionalInt(); o) {
    // use *o here
}

though the extra o check is kind of redundant.

---

This is one of those places where it'd be nice if optional<T> modeled a container of at most one element, so that you could do:

for (int value : GetOptionalInt()) {
    // possibly not entered
}

but we don't have that interface.

In order for this to work, there has to be a value for the unpacked value if it isn't there.

So

template<class T, class U>
std::pair< T, bool > unpack_value( std::optional<T> const& o, U&& u ) {
  return { o.value_or(std::forward<U>(u)), (bool)o } )
}

would do what you wanted.

But as an optional already returns if it is engaged in a bool context you really should just:

if (auto i = get_optional())

then use *i within the body.

...

Now if optional stated that operator* returned a reference, and that return value was defined but accessing it was not defined when it was not engaged, then you could write an Unpack method or function that doesn't require a default value.

As far as I am aware this is not true. And as it doesn't really add anything, I don't see why it should be true.