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.

Practice

1. What does it mean for a type in Go to implement an interface?

easy

A. The type can only have one method.

B. The type inherits from the interface.

C. The type must be explicitly declared as implementing the interface.

D. The type has all the methods declared in the interface.

Solution

1. Step 1: Understand interface implementation in Go

   In Go, a type implements an interface by having all the methods the interface declares.

2. Step 2: Compare options with Go's rules

   Go does not use inheritance or explicit declarations for interfaces; it uses method sets.

3. Final Answer:

   The type has all the methods declared in the interface. -> Option D

4. Quick Check:

   Interface implementation = method set match [OK]

Hint: Check if all interface methods exist in the type [OK]

Common Mistakes:

• Thinking Go uses inheritance for interfaces
• Believing explicit declaration is needed
• Confusing interfaces with structs

2. Which of the following is the correct way to declare that a type implements an interface in Go?

easy

A. type MyType struct implements MyInterface {}

B. type MyType struct { MyInterface }

C. No explicit declaration needed; implement all interface methods

D. func (m MyType) implements() MyInterface {}

Solution

1. Step 1: Recall Go interface implementation syntax

   Go does not require explicit declarations; a type implements an interface by having all its methods.

2. Step 2: Analyze each option

   Options A, B, and D show invalid or non-Go syntax; only C correctly states Go's implicit implementation.

3. Final Answer:

   No explicit declaration needed; implement all interface methods -> Option C

4. Quick Check:

   Implicit implementation = No explicit declaration needed; implement all interface methods [OK]

Hint: Remember: Go uses implicit interface implementation [OK]

Common Mistakes:

• Trying to declare implementation explicitly
• Using inheritance syntax from other languages
• Adding interface as a struct field

3. What will be the output of the following Go code?

package main
import "fmt"
type Speaker interface {
    Speak() string
}
type Dog struct {}
func (d Dog) Speak() string {
    return "Woof!"
}
func saySomething(s Speaker) {
    fmt.Println(s.Speak())
}
func main() {
    var d Dog
    saySomething(d)
}

medium

A. Woof!

B. Compilation error: Dog does not implement Speaker

C. Runtime error

D. No output

Solution

1. Step 1: Check if Dog implements Speaker

   Dog has a method Speak() string, so it implements Speaker interface.

2. Step 2: Trace the function call saySomething(d)

   saySomething calls s.Speak(), which returns "Woof!" and prints it.

3. Final Answer:

   Woof! -> Option A

4. Quick Check:

   Interface method call prints "Woof!" [OK]

Hint: Check if method signatures match interface exactly [OK]

Common Mistakes:

• Assuming explicit declaration is needed
• Confusing method receiver types
• Expecting runtime errors without reason

4. Identify the error in this Go code that tries to implement an interface:

package main
import "fmt"
type Runner interface {
    Run()
}
type Person struct {}
func (p *Person) Run() {
    fmt.Println("Running")
}
func main() {
    var r Runner
    p := Person{}
    r = p
    r.Run()
}

medium

A. Missing Run() method in Person

B. Cannot assign Person value to Runner interface because Run() has pointer receiver

C. Interface Runner must have a return type

D. No error, code runs fine

Solution

1. Step 1: Check method receiver type

   Run() is defined with pointer receiver (*Person), so only *Person implements Runner.

2. Step 2: Analyze assignment r = p

   Variable p is a Person value, not *Person pointer, so it does not implement Runner.

3. Final Answer:

   Cannot assign Person value to Runner interface because Run() has pointer receiver -> Option B

4. Quick Check:

   Pointer receiver method requires pointer type [OK]

Hint: Match method receiver type with assigned variable type [OK]

Common Mistakes:

• Ignoring pointer vs value receiver difference
• Assuming interface methods must return values
• Thinking interface assignment always works

5. Given these interfaces and types, which assignment is valid in Go?

type Reader interface {
    Read() string
}
type Writer interface {
    Write(string)
}
type ReadWriter interface {
    Reader
    Writer
}
type File struct {}
func (f *File) Read() string {
    return "data"
}
func (f *File) Write(s string) {
    // write data
}
func main() {
    var rw ReadWriter
    var f File
    // Which assignment is valid?
}

hard

A. rw = &f

B. rw = f

C. rw = File{} (without methods)

D. rw = nil

Solution

1. Step 1: Check if File implements ReadWriter

   File has Read() and Write(string) with pointer receivers (*File), so only *File implements ReadWriter; File value does not.

2. Step 2: Analyze assignment options

   rw = &f is valid (&f is *File). rw = f fails (File doesn't implement). C invalid. rw = nil compiles but doesn't hold usable ReadWriter.

3. Final Answer:

   rw = &f -> Option A

4. Quick Check:

   Pointer receiver requires pointer type [OK]

Hint: Pointer receiver methods mean only pointer type implements interface [OK]

Common Mistakes:

• Thinking value type implements pointer receiver interface
• Assigning nil to interface without initialization
• Confusing interface embedding with inheritance