How does Go implement code reuse without inheritance? Explain composition via struct embedding.

Go has no class hierarchy and no inheritance. Code reuse is achieved through composition: embedding one struct inside another promotes the embedded type's methods and fields to the outer type. This gives the outer type the embedded type's capabilities without any parent-child relationship.

// Base 'class' — a plain struct
type Animal struct {
    Name string
    Age  int
}
func (a Animal) Describe() string {
    return fmt.Sprintf("%s (age %d)", a.Name, a.Age)
}

// Composition: Dog embeds Animal
type Dog struct {
    Animal         // embedded — methods and fields promoted
    Breed string
}
func (d Dog) Sound() string { return "Woof" }

d := Dog{
    Animal: Animal{Name: "Rex", Age: 3},
    Breed:  "Labrador",
}
fmt.Println(d.Describe())  // promoted — same as d.Animal.Describe()
fmt.Println(d.Name)        // promoted field access
fmt.Println(d.Sound())     // Dog's own method

// Method overriding via shadowing
type PoliceDog struct {
    Dog
    BadgeNumber int
}
// PoliceDog can shadow Dog's promoted methods:
func (p PoliceDog) Sound() string {
    return "Woof! Police K9 #" + strconv.Itoa(p.BadgeNumber)
}

pd := PoliceDog{Dog: d, BadgeNumber: 42}
fmt.Println(pd.Sound())    // PoliceDog.Sound() — shadows Dog.Sound()
fmt.Println(pd.Dog.Sound()) // explicit Dog.Sound() — bypasses shadow
fmt.Println(pd.Describe())  // still promoted from Animal

Key distinction from inheritance: embedding is a mechanical promotion of methods, not an IS-A relationship. A PoliceDog is not a subtype of Dog — you cannot pass a PoliceDog where a Dog is expected (unless they share an interface). If both share an interface and implement all its methods, both can be used through that interface.

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