Quick Answer

Handle nil optionals safely.

Understanding the Issue

Swift requires optionals to be properly initialized before use, preventing null reference exceptions.

The Problem

This code demonstrates the issue:

Swift Error 
var name: String?
print(name!) // Crash if nil

The Solution

Here's the corrected code:

Swift Fixed 
// Solution 1: Optional binding
if let unwrappedName = name {
    print(unwrappedName)
}

// Solution 2: Nil coalescing
print(name ?? "default")

// Solution 3: Guard statement
guard let safeName = name else {
    print("Name is nil")
    return
}
print(safeName)

// Solution 4: Initialize properly
var name: String? = "Initial Value"

Key Takeaways

Always unwrap optionals safely using proper Swift idioms.