Type inference by the compiler in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Swift code using type inference?

Consider the following Swift code snippet. What will be printed?

Swift

let number = 42
let message = "The answer is \(number)"
print(message)

A 42

B The answer is number

C The answer is 42

D Error: Cannot infer type

💡 Hint

Look at how Swift infers the type of number and uses string interpolation.

Swift infers number as an Int. The string interpolation inserts the value 42 into the string, so the output is "The answer is 42".

❓ Predict Output

intermediate

What type does the compiler infer for this variable?

Given this Swift code, what type does the compiler infer for value?

Swift

let value = 3.14

A String

B Float

C Int

D Double

💡 Hint

Swift defaults to a certain floating-point type when no explicit type is given.

Swift infers value as a Double by default for floating-point literals.

🔧 Debug

advanced

Why does this Swift code cause a compile error?

Examine the code below. Why does the compiler produce an error?

Swift

let x = nil
print(x)

A print requires a non-optional argument

B Cannot infer type for nil without explicit type annotation

C nil is not allowed in Swift

D x is immutable and cannot be printed

💡 Hint

Think about what type nil represents and how Swift infers types.

Swift cannot infer the type of x because nil alone does not specify a type. You must provide a type annotation.

❓ Predict Output

advanced

What is the inferred type and output of this Swift closure?

Consider this Swift code snippet. What is the inferred type of increment and what will be printed?

Swift

let increment = { (number: Int) in number + 1 }
print(increment(5))

A Type: (Int) -> Int, Output: 6

B Type: (Int) -> Int, Output: 5

C Type: (Int) -> Void, Output: 6

D Type: () -> Int, Output: 6

💡 Hint

Look at the closure parameter and return expression.

The closure takes an Int and returns an Int by adding 1. Calling increment(5) returns 6.

🧠 Conceptual

expert

How does Swift's type inference handle mixed numeric literals in an expression?

What is the inferred type of result in the following Swift code?

let result = 5 + 3.0

A Double

B Int

C Float

D Error: Cannot mix Int and Double without casting

💡 Hint

Swift promotes numeric types to the more general type in mixed expressions.

Swift promotes the Int 5 to Double to match 3.0, so the result is inferred as Double.

Practice

1. What does type inference mean in Swift?

easy

A. The program runs slower because types are guessed.

B. You must always write the type explicitly.

C. The compiler automatically figures out the variable's type.

D. You cannot use variables without specifying types.

Solution

1. Step 1: Understand type inference concept

   Type inference means the compiler guesses the type of a variable based on the value assigned.

2. Step 2: Compare options with definition

   Only The compiler automatically figures out the variable's type. correctly describes automatic type detection by the compiler.

3. Final Answer:

   The compiler automatically figures out the variable's type. -> Option C

4. Quick Check:

   Type inference = automatic type detection [OK]

Hint: Type inference means compiler guesses type automatically [OK]

Common Mistakes:

• Thinking you must always write types explicitly
• Believing type inference slows down the program
• Confusing type inference with dynamic typing

2. Which of the following Swift variable declarations uses type inference correctly?

easy

A. let number = 10

B. let number: = 10

C. let number Int = 10

D. let number : Int 10

Solution

1. Step 1: Check syntax for variable declaration with type inference

   Type inference allows omitting the type, so let number = 10 is correct.

2. Step 2: Identify syntax errors in other options

   Options B, C, and D have syntax errors: missing type name, wrong order, or missing assignment operator.

3. Final Answer:

   let number = 10 -> Option A

4. Quick Check:

   Correct syntax with inference = let number = 10 [OK]

Hint: Look for correct Swift syntax without explicit type [OK]

Common Mistakes:

• Adding colon without type name
• Writing type before variable name incorrectly
• Omitting assignment operator

3. What is the type of variable value in this Swift code?

let value = 3.14

medium

A. Double

B. Int

C. String

D. Float

Solution

1. Step 1: Identify the literal type assigned

   The number 3.14 is a decimal number, which Swift treats as Double by default.

2. Step 2: Confirm Swift default for floating-point literals

   Swift infers floating-point literals as Double unless specified otherwise.

3. Final Answer:

   Double -> Option A

4. Quick Check:

   3.14 literal = Double type [OK]

Hint: Decimal numbers default to Double in Swift [OK]

Common Mistakes:

• Assuming 3.14 is Float by default
• Confusing Int with floating-point types
• Thinking literals default to String

4. Identify the error in this Swift code snippet:

var name = "Alice"
name = 42

medium

A. No error, type inference allows changing type.

B. Error: Cannot assign Int to a variable inferred as String.

C. Error: Missing semicolon after first line.

D. Error: Variable must be declared with explicit type.

Solution

1. Step 1: Check initial type inferred for variable

   Variable name is assigned a String "Alice", so compiler infers type String.

2. Step 2: Analyze reassignment with different type

   Assigning an Int (42) to a String variable causes a type mismatch error.

3. Final Answer:

   Error: Cannot assign Int to a variable inferred as String. -> Option B

4. Quick Check:

   Type mismatch on reassignment = Error [OK]

Hint: Variable type fixed by first assignment, cannot change later [OK]

Common Mistakes:

• Thinking type inference allows changing variable type
• Assuming semicolons are required in Swift
• Believing explicit type declaration is always needed

5. Given this Swift code, what is the type of result?

let numbers = [1, 2, 3]
let result = numbers.map { $0 * 2 }

hard

A. Array<Any>

B. Int

C. [Double]

D. [Int]

Solution

1. Step 1: Understand the type of numbers

   Array numbers contains Int values, so its type is [Int].

2. Step 2: Analyze the map operation and type inference

   The map multiplies each Int by 2, producing another Int for each element, so the result is an array of Ints.

3. Step 3: Confirm inferred type of result

   Since map returns an array of transformed elements, result is inferred as [Int].

4. Final Answer:

   [Int] -> Option D

5. Quick Check:

   map on [Int] with Int output = [Int] [OK]

Hint: map returns array of same type as transformed elements [OK]

Common Mistakes:

• Thinking map returns a single Int
• Assuming map changes type to Double automatically
• Believing result is Array<Any> without type inference