Convert optional substring to string in Swift

Question

Apple's documentation and many articles suggest you can convert a substring (string.subsequence) to a string just by calling String(substring)and in fact this works.

let str = "Hello world"
let hello = str.prefix(5) //create substr
//CREATE STRING FROM SUBSTRING
let hellostr = String(hello) //works

However when the substring is an optional as in the following,

 let str = Optional("Hello world")
 let hello = str?.prefix(5) //create substr
 //CREATE STRING FROM OPTIONAL SUBSTRING
 let hellostr = String(hello) //gives error

I am getting the error

"No exact matches in call to initializer "

What could be causing this error?

Thanks for any suggestions.

The method is init(_ substring: Substring), not init(_ substring: Substring?). So String(hello ?? Substring()) should do the trick. Also the comments about the code " //gives error " is false and misleading (I guess wrong copy/paste) — Larme
– Larme, Commented Jul 5, 2024 at 11:51
What is the goal to create a string from nil? Practically there is none therefore an init method is pointless. — vadian
– vadian, Commented Jul 5, 2024 at 11:59

David Pasztor · Accepted Answer · 2024-07-05 13:48:33Z

4

You need to unwrap the Optional Substring as you cannot directly create a String from an Optional Substring.

You need to decide how you want to handle nil values for hello and based on that, choose the appropriate option.

Here's a few options:

Use Optional.map to create a non-optional String from the Substring in case the optional had a non-nil value and if the optional was nil, assign nil to the result/

// hellostr will be nil if hello was nil, otherwise it will have a value
// The type of hellostr is Optional<String>
let hellostr = hello.map(String.init)

If you always want hellostr to have a value, you need to provide a default value either to the return value of Optional.map
```
let hellostr = hello.map(String.init) ?? ""
```
or (as Martin R pointed out in comments) to the Optional<Substring> passed to String.init:
```
let hellostr = String(hello ?? "")
```

You can also use optional binding to replace Optional.map with more verbose syntax

let hellostr: String
if let hello = hello {
  hellostr = String(hello)
} else {
  hellostr = ""
}

edited Jul 5, 2024 at 13:48

answered Jul 5, 2024 at 11:50

David Pasztor

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1 Comment

Martin R Over a year ago

An alternative for #2 is let hellostr = String(hello ?? "")

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