Why is count ambiguous here?

Using Swift
1 
 let x = [1: ["a", "b", "c", "d"],
      2: ["e", "f", "g"],
      3: ["h", "i"]]
 
 let total = x.reduce(0) {&$0 += $1.value.count}
2

Wow, that's terse. Try breaking that expression up, my guess is this is actually not the right error, but the type checker is getting confused.

3

The error message is misleading. The closure must return the accumulated value:

let total = x.reduce(0) { $0 + $1.value.count }

Or perhaps you meant to use reduce(into:_:) ? Here the accumulated value is passed as an inout argument to the closure and can be updated:

let total1 = x.reduce(into: 0) { $0 += $1.value.count }
6 Likes
4

This is simply a diagnostics engine bug. Believe it or not, typing something on the next line fixes the error message:

_ = x.reduce(0) { f, s in
  
  &f += s.value.count // Ambiguous reference to member 'count'
}

_ = x.reduce(0) { f, s in
  
  &f += s.value.count // Cannot pass immutable value as inout argument: 'f' is a 'let' constant
  "ffff"
}

I'm going to file this if you don't mind. UDP SR-7543

3 Likes
5

Thanks @Martin. That's what I meant to do.