How can I concatenate multiple optional strings in swift 3.0? - string

Question

I am trying to concatenate multiple strings in swift 3:
var a:String? = "a"
var b:String? = "b"
var c:String? = "c"
var d:String? = a! + b! + c!
When compiling I get the following error:
error: cannot convert value of type 'String' to specified type 'String?'
var d:String? = a! + b! + c!
~~~~~~~~^~~~
This used to work in swift 2. I am not sure why it doesn't work anymore.

Answer 1

Bug report filed by OP:
SR-1122: Failure to typecheck chain of binary operators on force-unwrapped values
Which has been resolved (fix commited to master Jan 3 2017), and should hence no longer be an issue in upcoming Swift 3.1.
This seems to be a bug (not present in Swift 2.2, only 3.0) associated with the case of:
Using the forced unwrapping operator (!) for at least 3 terms in an expression (tested using at least 2 basic operators, e.g. + or -).
For some reason, given the above, Swift messes up type inference of the expression (specifically, for the x! terms themselves, in the expression).
For all the examples below, let:
let a: String? = "a"
let b: String? = "b"
let c: String? = "c"
Bug present:
// example 1
a! + b! + c!
/* error: ambiguous reference to member '+' */
// example 2
var d: String = a! + b! + c!
/* error: ambiguous reference to member '+' */
// example 3
var d: String? = a! + b! + c!
/* error: cannot convert value of type 'String'
to specified type 'String?' */
// example 4
var d: String?
d = a! + b! + c!
/* error: cannot assign value of type 'String'
to specified type 'String?' */
// example 5 (not just for type String and '+' operator)
let a: Int? = 1
let b: Int? = 2
let c: Int? = 3
var d: Int? = a! + b! + c!
/* error: cannot convert value of type 'Int'
to specified type 'Int?' */
var e: Int? = a! - b! - c! // same error
Bug not present:
/* example 1 */
var d: String? = a! + b!
/* example 2 */
let aa = a!
let bb = b!
let cc = c!
var d: String? = aa + bb + cc
var e: String = aa + bb + cc
/* example 3 */
var d: String? = String(a!) + String(b!) + String(c!)
However as this is Swift 3.0-dev, I'm uncertain if this is really a "bug", as well as what's the policy w.r.t. reporting "bugs" in a not-yet-production version of code, but possibly you should file radar for this, just in case.
As for answering your question as how to circumvent this issue:
use e.g. intermediate variables as in Bug not present: Example 2 above,
or explicitly tell Swift all terms in the 3-term expression are strings, as in Bug not present: Example 3 above,
or, better yet, use safe unwrapping of your optional, e.g. using optional binding:
var d: String? = nil
if let a = a, b = b, c = c {
d = a + b + c
} /* if any of a, b or c are 'nil', d will remain as 'nil';
otherwise, the concenation of their unwrapped values */

Answer 2

Swift 3
let q: String? = "Hello"
let w: String? = "World"
let r: String? = "!"
var array = [q, w, r]
print(array.flatMap { $0 }.reduce("", {$0 + $1}))
// HelloWorld!
let q: String? = "Hello"
let w: String? = nil
let r: String? = "!"
var array = [q, w, r]
print(array.flatMap { $0 }.reduce("", {$0 + $1}))
// Hello!

Answer 3

func getSingleValue(_ value: String?..., seperator: String = " ") -> String? {
return value.reduce("") {
($0) + seperator + ($1 ?? "")
}.trimmingCharacters(in: CharacterSet(charactersIn: seperator) )
}

Answer 4

let val: String? = "nil"
val.flatMap({(str: String) -> String? in
return str + "value"
})

Answer 5

var a:String? = "a"
var b:String? = "b"
var c:String? = "c"
var d:String? = ""
let arr = [a,b,c]
arr.compactMap { $0 }.joined(separator: " ")
compactMap be used to filter out nil values from flattened arrays