Go Interfaces Value & pointer receivers

Links:

• 103 Golang Index
• Go - Interfaces

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Interface : Value and pointer receivers

• Example with value receiver

  type myStruct struct {
  }
  
  func (p1 myStruct) print1(a int) {
      fmt.Println(a)
  }
  
  func (p2 myStruct) print2(a string) {
      fmt.Println(a)
  }
  
  type myInterface interface {
      print1(int)
      print2(string)
  }
  
  func main() {
      var ms myInterface = myStruct{} // value type for implementing an interface
      fmt.Println(ms)
  }
  

When implementing an interface if you use a value type, all the methods should have value receivers. If we are implementing an interface with a pointer then we should just have the methods there.

Even if none of the functions were pointer receivers, using a pointer would have worked.

func (p2 *myStruct) print2(a string) {
    fmt.Println(a)
}
// this would give an error since we have a pointer receiver

// to fix this we can do
var ms myInterface = &myStruct{}

Another Example

In the below example Vertex (the value type) doesn't implement Abser because the Abs method is defined only on *Vertex (the pointer type).

type Abser interface {
    Abs() float64
}

func main() {
    var a Abser
    f := MyFloat(-math.Sqrt2)
    v := Vertex{3, 4}

    a = f  // a MyFloat implements Abser
    a = &v // a *Vertex implements Abser

    // In the following line, v is a Vertex (not *Vertex)
    // and does NOT implement Abser.
    a = v

    fmt.Println(a.Abs())
}

type MyFloat float64

func (f MyFloat) Abs() float64 {
    if f < 0 {
        return float64(-f)
    }
    return float64(f)
}

type Vertex struct {
    X, Y float64
}

func (v *Vertex) Abs() float64 {
    return math.Sqrt(v.X*v.X + v.Y*v.Y)
}

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Last updated: 2022-06-29