Implementing interfaces in Go - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare that type Dog implements the Speaker interface.

Go

type Speaker interface {
    Speak() string
}

type Dog struct {}

func (d Dog) [1]() string {
    return "Woof!"
}

Drag options to blanks, or click blank then click option'

ASound

BTalk

CSay

DSpeak

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different method name like Talk or Say.

Not matching the method signature exactly.

The method name must exactly match the interface method name Speak to implement the interface.

2fill in blank

medium

Complete the code to create a variable of type Speaker and assign a Dog value to it.

Go

var s Speaker = [1]{}

Drag options to blanks, or click blank then click option'

ADog

BFish

CCat

DBird

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning a type that does not implement Speaker.

Using a pointer type when not required.

Dog implements Speaker, so we can assign a Dog value to a Speaker variable.

3fill in blank

hard

Fix the error in the method receiver to correctly implement the Speaker interface.

Go

func (d *Dog) [1]() string {
    return "Woof!"
}

Drag options to blanks, or click blank then click option'

ASpeak

BTalk

CSay

DSound

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different method name.

Not matching the interface method signature.

The method name must be Speak to implement the Speaker interface.

4fill in blank

hard

Fill both blanks to define a Cat type that implements Speaker with a pointer receiver.

Go

type Cat struct {}

func ([1] *Cat) [2]() string {
    return "Meow!"
}

Drag options to blanks, or click blank then click option'

Ac

BCat

CSpeak

DTalk

Attempts:

3 left

💡 Hint

Common Mistakes

Using the type name as receiver variable.

Wrong method name.

The receiver variable is usually a short name like c, and the method name must be Speak to implement the interface.

5fill in blank

hard

Fill all three blanks to create a function that accepts a Speaker and calls its Speak method.

Go

func saySomething([1] [2]) string {
    return [1].[3]()
}

Drag options to blanks, or click blank then click option'

As

BSpeaker

CSpeak

DDog

Attempts:

3 left

💡 Hint

Common Mistakes

Using a concrete type instead of the interface.

Calling a method not defined in the interface.

The function parameter name is s, its type is Speaker, and we call s.Speak() to get the string.