[swift-users] Ambiguous parsing of `return`

[Daniel Dunbar]

Did you try an example? Swift warns about this for this very reason:
```
$ cat x.swift 
func f0() {}

func f1() {
  return
  f0()
}

$ swiftc -c x.swift 
x.swift:5:3: warning: expression following 'return' is treated as an argument of the 'return'
  f0()
  ^
```

 - Daniel

> On Jun 6, 2016, at 10:00 AM, Jens Alfke via swift-users <swift-users at swift.org> wrote:
> 
> Someone on another forum (not directly related to Swift) just mentioned running into a confusing situation where they had added an early `return` to a method for testing purposes, in order to disable the code following it:
> 
> func myFunc() {
>         // some code
>         return
>         cache.removeAll()
>         // more code that is now skipped.
> }
> 
> Unexpectedly, the line following the `return` still got executed, so "I couldn't figure out why my cache kept getting zapped.”
> 
> Turns out the Swift parser is interpreting this as `return cache.removeAll()`, which works because that expression returns void, which matches the function’s return type.
> 
> This seems like a case where the parser is playing by the rules, but the result is not what a human would expect. It would be better for a `return` on a line by itself to be parsed as a complete statement, without continuing to the next line. Is this already a known issue?
> 
> —Jens
> 
> PS: I’m sure someone will point out that adding an early return like this is sub-optimal, and the compiler could warn that the code following is unreachable. Which is true, and I use comments to disable code in situations like this. But I’m sure this developer’s not the only one who adds `return` instead.
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