Golang 运算符重载

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)
}

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
}

A

A

Add()

Add()

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

Add()

Add()

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

a.Add()(&a).Add()

a.Add()(&a).Add()

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