Type conversion is always explicit in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of explicit type conversion in Swift

What is the output of this Swift code snippet?

Swift

let intVal: Int = 10
let doubleVal: Double = 3.5
let result = Double(intVal) + doubleVal
print(result)

A 13.5

B 13

C Type error at compile time

D 10.0

💡 Hint

Remember that Swift requires explicit conversion between Int and Double types.

The integer 10 is explicitly converted to Double before addition, so 10.0 + 3.5 equals 13.5.

❓ Predict Output

intermediate

Result of forced type conversion with optional

What will be printed by this Swift code?

Swift

let str = "123"
let num = Int(str)!
print(num + 1)

A 124

B 1231

C Type error at compile time

D Runtime error: nil unwrapping

💡 Hint

The string contains only digits, so conversion to Int succeeds.

Int(str) converts the string "123" to integer 123. Forced unwrapping (!) succeeds, so 123 + 1 equals 124.

❓ Predict Output

advanced

What error occurs with implicit conversion?

What error does this Swift code produce?

Swift

let a: Int = 5
let b: Double = 2.0
let c = a + b
print(c)

A 7.0

B Type error: binary operator '+' cannot be applied to operands of type 'Int' and 'Double'

C 5 + 2 = 7

D Runtime error: cannot add Int and Double

💡 Hint

Swift does not allow implicit conversion between Int and Double in arithmetic operations.

Swift requires explicit conversion; adding Int and Double without conversion causes a compile-time type error.

❓ Predict Output

advanced

Output of converting Double to Int explicitly

What is the output of this Swift code?

Swift

let pi: Double = 3.99
let intPi = Int(pi)
print(intPi)

A 4

B Type error at compile time

C 3.99

D 3

💡 Hint

Converting Double to Int truncates the decimal part.

Int(pi) truncates the decimal part, so 3.99 becomes 3.

🧠 Conceptual

expert

Why is explicit type conversion required in Swift?

Which of the following best explains why Swift requires explicit type conversion between numeric types?

A To improve runtime performance by avoiding automatic conversions

B Because Swift does not support numeric operations between different types

C To prevent unexpected data loss or bugs by making conversions clear and intentional

D Because implicit conversions are handled automatically by the compiler

💡 Hint

Think about safety and clarity in code.

Explicit conversions make the programmer aware of possible data loss or changes, preventing subtle bugs.

Practice

1. In Swift, why must type conversion be explicit?
let x = 5 and let y = 3.2 are different types. What happens if you try to add x + y without conversion?

easy

A. Swift automatically converts x to a Double and adds them.

B. It causes a compile-time error because Swift does not convert types automatically.

C. The program runs but the result is always an integer.

D. Swift converts y to an Int silently.

Solution

1. Step 1: Understand Swift's type safety

   Swift requires explicit type conversion to avoid unexpected behavior.

2. Step 2: Check adding Int and Double without conversion

   Trying to add Int and Double directly causes a compile-time error.

3. Final Answer:

   It causes a compile-time error because Swift does not convert types automatically. -> Option B

4. Quick Check:

   Int + Double without conversion causes compile error [OK]

Hint: Remember: Swift never converts types automatically [OK]

Common Mistakes:

• Assuming Swift converts Int to Double automatically
• Thinking the result is always an Int
• Believing silent conversion happens

2. Which of the following is the correct way to convert an Int num = 10 to a Double in Swift?

easy

A. let d = convert(num, to: Double)

B. let d = num as Double

C. let d = Double(num)

D. let d = num.toDouble()

Solution

1. Step 1: Recall Swift's explicit conversion syntax

   Swift uses the target type name as a function to convert values.

2. Step 2: Identify correct syntax for Int to Double

   The correct syntax is Double(num). Other options are invalid in Swift.

3. Final Answer:

   let d = Double(num) -> Option C

4. Quick Check:

   Target type as function: Double(num) [OK]

Hint: Use target type name with parentheses for conversion [OK]

Common Mistakes:

• Using 'as' for numeric conversion
• Calling non-existent methods like toDouble()
• Using custom convert functions not in Swift

3. What is the output of this Swift code?

let a: Int = 7
let b: Double = 3.5
let c = Double(a) + b
print(c)

medium

A. 10.5

B. 10

C. 7.0

D. Error: Cannot add Int and Double

Solution

1. Step 1: Convert Int to Double explicitly

   a is Int 7, converted to Double 7.0 using Double(a).

2. Step 2: Add converted Double and b

   Adding 7.0 + 3.5 results in 10.5.

3. Final Answer:

   10.5 -> Option A

4. Quick Check:

   7.0 + 3.5 = 10.5 [OK]

Hint: Convert Int to Double before adding [OK]

Common Mistakes:

• Forgetting to convert Int to Double
• Expecting integer addition result
• Assuming implicit conversion

4. Identify the error in this Swift code snippet:

let str = "123"
let num = Int(str)
let result = num + 10
print(result)

medium

A. Missing parentheses in Int conversion

B. String cannot be converted to Int

C. No error, code runs fine

D. Cannot add Int? and Int without unwrapping

Solution

1. Step 1: Understand Int initializer returns optional

   Int(str) returns Int? because conversion might fail.

2. Step 2: Adding optional Int and Int causes error

   You cannot add Int? and Int directly without unwrapping.

3. Final Answer:

   Cannot add Int? and Int without unwrapping -> Option D

4. Quick Check:

   Int(str) returns Int?, must unwrap before adding [OK]

Hint: Unwrap optionals before arithmetic operations [OK]

Common Mistakes:

• Ignoring optional unwrapping
• Assuming Int(str) returns Int directly
• Trying to add optional and non-optional

5. You have an array of strings representing numbers: let arr = ["1", "2", "three", "4"]. You want to convert them to Ints and sum only the valid numbers. Which Swift code correctly does this?

hard

A. let sum = arr.compactMap { Int($0) }.reduce(0, +)

B. let sum = arr.map { Int($0) }.reduce(0, +)

C. let sum = arr.reduce(0) { $0 + Int($1)! }

D. let sum = arr.reduce(0) { $0 + Int($1) }

Solution

1. Step 1: Use compactMap to convert and filter valid Ints

   compactMap converts strings to Ints and removes nil values safely.

2. Step 2: Use reduce to sum the valid Ints

   reduce(0, +) sums all Ints in the array.

3. Final Answer:

   let sum = arr.compactMap { Int($0) }.reduce(0, +) -> Option A

4. Quick Check:

   compactMap filters nils, then sum [OK]

Hint: Use compactMap to safely convert and filter nils [OK]

Common Mistakes:

• Using map which keeps optionals causing errors
• Force unwrapping without checking validity
• Adding optionals directly without unwrapping