Default parameter values in Swift - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of function with default parameters

What is the output of this Swift code when calling greet() without any arguments?

Swift

func greet(name: String = "Friend", greeting: String = "Hello") {
    print("\(greeting), \(name)!")
}
greet()

A Error: Missing arguments

B Hello, !

C Friend, Hello!

D Hello, Friend!

💡 Hint

Check the default values assigned to the parameters in the function definition.

The function greet has default values for both parameters. Calling greet() uses those defaults, printing "Hello, Friend!".

❓ Predict Output

intermediate

Function call with one default parameter overridden

What will be printed when calling greet(name: "Alice") with the function below?

Swift

func greet(name: String = "Friend", greeting: String = "Hello") {
    print("\(greeting), \(name)!")
}
greet(name: "Alice")

A Hello, Alice!

B Alice, Hello!

C Error: Missing argument for greeting

D Hello, Friend!

💡 Hint

Only the name parameter is provided; the other uses its default.

The name parameter is set to "Alice" explicitly, while greeting uses its default "Hello".

❓ Predict Output

advanced

Output with default parameters and argument labels

What is the output of this Swift code?

Swift

func calculateArea(width: Int = 10, height: Int = 5) -> Int {
    return width * height
}
print(calculateArea(height: 3))

A 50

B Error: Missing argument for width

C 30

D 15

💡 Hint

The width parameter uses its default value.

The call provides height: 3 but omits width, so width defaults to 10. The area is 10 * 3 = 30.

❓ Predict Output

advanced

Default parameter with mutable variable

What will be printed after this code runs?

Swift

var count = 5
func increment(by amount: Int = count) -> Int {
    return amount + 1
}
count = 10
print(increment())

A Error: Cannot use variable 'count' as default parameter

B 11

C 5

D 6

💡 Hint

Swift requires default parameter values to be compile-time constants; variables are not allowed.

Swift does not allow using variables as default parameter values because they must be compile-time constants. This code will not compile.

🧠 Conceptual

expert

Why default parameter values must be compile-time constants

Why does Swift require default parameter values to be compile-time constants or literals?

A Because Swift does not support functions with optional parameters

B Because default values are evaluated once at compile time and cannot depend on runtime variables

C Because default values are ignored during function calls

D Because default values must be mutable to change during execution

💡 Hint

Think about when the default value is set and how it relates to program execution.

Swift evaluates default parameter values at compile time to ensure consistent behavior and avoid runtime errors from changing values.