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

Question

I'm writing some code using std::optional's and am wondering if C++17's 'if statements with initializers' will be able to help unpack values?

std::optional<int> optionalInt = GetOptionalInt();

I'm making up the function Unpack here:

if( auto [value, has_value] = optionalInt.Unpack(); has_value )
{
    // Use value here.
}

But, my question is. Will C++17 'if statement with initializer' help here? If so, how would it be coded?

Update, this is actually mainly an issue when using optional which is extremely easy to misuse because the optional and *optional both return bools and you don't get any compiler warning when somebody trys to access the value and forgets the *.

Answer 1

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.