分配给变量或立即返回它不起作用

Question

If I do this:

func myFunc() -> NSRange{
    var range = s.rangeOfString(searchText, options: .CaseInsensitiveSearch)
    return s.rangeOfString(searchText, options: .CaseInsensitiveSearch)
}

I get an error that says there's extra arguments in the second call. But they are exactly the same!

On the other hand, this works fine:

func myFunc() -> NSRange{
    var range = s.rangeOfString(searchText, options: .CaseInsensitiveSearch)
    return range
}

My question is implicit. So here it is: why swift compiler doesn't get satisfied with the first option and instead it makes me assign the return of the function to a variable first?

If that works for you, here's my full code:

var a: [MyObject] = []
array = a.filter{
                var b = ($0 as MyObject).molecule.name
                var s: NSString = b as NSString
                var range = s.rangeOfString(searchText, options: .CaseInsensitiveSearch)
                return s.rangeOfString(searchText, options: .CaseInsensitiveSearch)
                return range.location != -1
            }

Answer 1

The problem is different than what Swift reports. The error reporting in Swift can be pretty strange at times.

The real problem is that filter expects a boolean value to be returned, representing if that item should be included. But rangeOfString returns an NSRange .

So if you return an expression that yields a boolean value, it works fine.

return s.rangeOfString(
  searchText,
  options: .CaseInsensitiveSearch
).location != NSNotFound

---

Which is why your first example works fine. You say this function returns an NSRange and then you return an NSRange so it works!

func myFunc() -> NSRange{
    var range = s.rangeOfString(searchText, options: .CaseInsensitiveSearch)
    return s.rangeOfString(searchText, options: .CaseInsensitiveSearch)
}