Quick Answer

Fix optional binding syntax.

Understanding the Issue

Swift requires optional types for if-let and guard-let bindings to safely unwrap values.

The Problem

This code demonstrates the issue:

Swift Error 
let num = 42
if let unwrapped = num { // Error
    print(unwrapped)
}

The Solution

Here's the corrected code:

Swift Fixed 
// Solution 1: Make the value optional
let optionalNum: Int? = 42
if let unwrapped = optionalNum {
    print(unwrapped)
}

// Solution 2: Use non-optional directly
let num = 42
print(num)

// Solution 3: Optional cast
let anyValue: Any = 42
if let number = anyValue as? Int {
    print(number)
}

// Solution 4: Provide default value
print(optionalNum ?? 0)

Key Takeaways

Only use optional binding with actual optional types.