Golang 运算符重载 [英] Golang Operator Overloading

问题描述

我知道 golang 不提供运算符重载，因为它认为它会增加复杂性.

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

所以我想直接为结构实现它.

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

从上面的代码中，AddValue 可以按照我的意愿工作.但是，我唯一担心的是它是传值，因此我每次都必须返回新添加的值.

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.

有没有其他更好的方法，以避免返回汇总的变量.

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

推荐答案

是的，使用指针接收器:

Yes, use pointer receiver:

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

通过使用指针接收器，A 类型的值的地址将被传递，因此如果您修改指向的对象，您不必返回它，您将修改原始"对象而不是副本.

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.

您也可以简单地将其命名为 Add().你也可以让它的参数成为一个指针(为了一致性):

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
}

所以使用它:

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

x.Add(y)

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

注意事项

请注意，即使您现在有一个指针接收器，如果非指针值可寻址，您仍然可以在非指针值上调用您的 Add() 方法，因此例如以下也有效:

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() 是 (&a).Add() 的简写.在 Go Playground 上试试这些.

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

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