Type Switch Operator

[DavePearce]

The type switch operator is inspired by the ? operator in Rust. Some points:

• The ? operator is inextricably linked with std::error::Error and is used for error propagation.
• For Whiley, a more general version of it is desirable.

In particular, since we have union types in Whiley, the ? operator should work with them. For example:

function test() -> (int|bool r):
   int x = other()?
   return x + 1

function other() -> (int|bool r):
   1

What's happening here is that, by exploiting forward type information from x the ? operator attempts to resolve by pulling out the desired item, otherwise return what is left from the enclosing function. For that to type check, what's left must fit into the return type.

This then allows us to encode exceptional control flow roughly as is done in Rust.

[DavePearce]

The ? operator is actually a bit more general than I gave it credit for above. For example, the following is fine:

pub fn test() -> Option<u32> {
   let x = other()?;
   Some(x + 1)
}

pub fn other() -> Option<u32> { Some(1) }

However, the following does not work and the compiler complains that ? needs either an Option or a Result.

enum Union<X,Y> { Left(X), Right(Y) }

pub fn test() -> Union<u32,bool> {
   let x = other()?;
   Union::Left(x + 1)
}

pub fn other() -> Union<u32,bool> { Union::Left(1) }