Golang Operator Overloading

问题

I understand that golang does not provide operator overloading, as it believe that it is increasing the complexity.

So I want to implement that for structures directly.

package main

import "fmt"

type A struct {
    value1 int
    value2 int
}

func (a A) AddValue(v A) A {
    a.value1 += v.value1
    a.value2 += v.value2
    return a
}


func main() {
    x, z := A{1, 2}, A{1, 2}
    y := A{3, 4}

    x = x.AddValue(y)

    z.value1 += y.value1
    z.value2 += y.value2

    fmt.Println(x)
    fmt.Println(z)
}

https://play.golang.org/p/1U8omyF8-V

From the above code, the AddValue works as I want to. However, my only concern is that it is a pass by value and hence I have to return the newly added value everytime.

Is there any other better method, in order to avoid returning the summed up variable.

回答1:

Yes, use pointer receiver:

func (a *A) AddValue(v A) {
    a.value1 += v.value1
    a.value2 += v.value2
}

By using a pointer receiver, the address of a value of type A will be passed, and therefore if you modify the pointed object, you don't have to return it, you will modify the "original" object and not a copy.

You could also simply name it Add(). And you could also make its argument a pointer (for consistency):

func (a *A) Add(v *A) {
    a.value1 += v.value1
    a.value2 += v.value2
}

And so using it:

x, y := &A{1, 2}, &A{3, 4}

x.Add(y)

fmt.Println(x)  // Prints &{4 6}

Notes

Note that even though you now have a pointer receiver, you can still call your Add() method on non-pointer values if they are addressable, so for example the following also works:

a, b := A{1, 2}, A{3, 4}
a.Add(&b)
fmt.Println(a)

a.Add() is a shorthand for (&a).Add(). Try these on the Go Playground.

来源：https://stackoverflow.com/questions/33040495/golang-operator-overloading

标签

methods

struct

operator-overloading