Implementing interfaces in Go - Step-by-Step Execution

Concept Flow - Implementing interfaces

Define Interface

↓

Define Struct

↓

Implement Interface Methods on Struct

↓

Create Struct Instance

↓

Assign to Interface Variable

↓

Call Interface Methods

↓

Use Polymorphism

This flow shows how a struct implements an interface by defining required methods, then using the struct through the interface type.

Execution Sample

Go

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

This code defines an interface Speaker and a struct Dog that implements it, then calls Speak through the interface.

Execution Table

Step	Action	Evaluation	Result
1	Define interface Speaker with method Speak()	Interface created	Speaker interface ready
2	Define struct Dog	Struct created	Dog struct ready
3	Implement Speak() method on Dog	Method attached	Dog now implements Speaker
4	Create Dog instance and assign to Speaker variable s	Dog{} assigned to s	s holds Dog as Speaker
5	Call s.Speak()	Calls Dog.Speak()	"Woof!" returned and printed
6	Program ends	No more code	Execution stops

💡 Program ends after printing "Woof!"

Variable Tracker

Variable	Start	After Step 4	After Step 5	Final
s	nil	Dog{} as Speaker	Dog{} as Speaker	Dog{} as Speaker

Key Moments - 3 Insights

Why can we assign Dog{} to a variable of type Speaker without explicit casting?

Because Dog implements all methods of Speaker (Speak()), Go allows implicit assignment as shown in step 4 of the execution_table.

What happens if Dog does not implement Speak() method?

The assignment in step 4 would fail with a compile-time error, because Dog wouldn't satisfy the Speaker interface.

When calling s.Speak(), which method runs?

The Dog's Speak() method runs, as s holds a Dog instance implementing Speaker, shown in step 5.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of variable s after step 4?

A Speaker interface without value

B nil

C Dog{} as Speaker

D Dog struct without interface

💡 Hint

Check variable_tracker row for s after step 4

At which step does Dog become recognized as implementing Speaker?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at execution_table step 3 where Speak() method is attached

If Dog did not have Speak() method, what would happen at step 4?

A Compile-time error occurs

B Program runs but panics at runtime

C Assignment succeeds with nil value

D Dog is assigned but Speak() returns empty string

💡 Hint

Refer to key_moments about missing method causing assignment failure

Concept Snapshot

Go interfaces define method sets.
A struct implements an interface by having those methods.
No explicit declaration needed.
Assign struct instance to interface variable.
Call methods via interface for polymorphism.

Full Transcript

This example shows how in Go, interfaces are implemented by structs automatically when the struct has all the methods the interface requires. We define an interface Speaker with method Speak(). Then we define a struct Dog and add a Speak() method to it. Because Dog has Speak(), it implements Speaker. We create a Dog instance and assign it to a variable of type Speaker. When we call Speak() on that variable, Dog's Speak() runs and returns "Woof!". This demonstrates Go's interface implementation and polymorphism.