Swift 6.2 introduces a subtle but powerful enhancement through SE-0477: the ability to specify default values directly within string interpolations for optionals. It’s one of those changes that may look trivial at first glance — but under the surface, it solves a long-standing pain point and brings elegance to previously clunky code.

The Basic Idea

Let’s start with the core of this change. Suppose you have an optional string:

var name: String?

Previously, to provide a fallback value inside a string, you’d write:

print("Hello, \(name ?? "No name")")

Now, with Swift 6.2, you can do this instead:

print("Hello, \(name, default: "No name")")

A cleaner syntax, sure — but is it really better than nil coalescing?

Looks Redundant? Not Quite

At first glance, this change might seem cosmetic. After all, the ?? operator already does a great job, right? Why introduce a new syntax that feels longer and arguably more verbose?

Here’s the twist: the new interpolation syntax works even when the types don’t match — something that ?? simply can’t do.

Let’s take a look at this:

var count: Int?
print("Count: \(count ?? .zero)") // OK

Now try this:

print("Count: \(count ?? "Unknown")") // Compiler error

The compiler throws a fit, because count is an optional Int, and "Unknown" is a String. The ?? operator expects both sides to be of the same type.

But in Swift 6.2:

print("Count: \(count, default: "Unknown")") // Works perfectly

The default value is used directly as a string in the interpolation, regardless of the type of the optional. That subtle shift makes a world of difference in some scenarios.

When This Really Shines: Complex Interpolations

Let’s consider a more practical, real-world example — say, displaying user profile data:

struct User {
    var username: String?
    var email: String?
    var phoneNumber: String?
    var age: Int?
}

let user = User(
    username: nil,
    email: "jane@example.com",
    phoneNumber: nil,
    age: nil
)

print("""
User Info:
- Username: \(user.username, default: "Guest")
- Email: \(user.email, default: "Not provided")
- Phone: \(user.phoneNumber, default: "Not available")
- Age: \(user.age, default: "Unknown")
""")

Here, using ?? would either require type casting, temporary variables, or verbose conditional logic. The new interpolation syntax keeps it readable and clean — especially when working with many optional fields of varying types.

Another Useful Example: Logging and Debugging

func logEvent(name: String?, duration: Double?) {
    print("Event '\(name, default: "Unnamed")' ran for \(duration, default: "an unknown amount of time")")
}

logEvent(name: nil, duration: nil)
// Output: Event 'Unnamed' ran for an unknown amount of time

Before Swift 6.2, you’d have to convert numbers to strings manually or extract and preprocess values just to avoid crashing the interpolation. This update lets you handle missing data more gracefully, right at the point of use.

Final Thoughts

Yes, it may look like a redundant feature — until you hit a case where optional values of different types need to be interpolated directly into a string. That’s where this small enhancement becomes a big win for code clarity and developer ergonomics.

Clean, safe, and expressive — just like Swift should be.