英文:
Golang Operator Overloading
问题
我理解到，Golang不提供运算符重载，因为它认为这会增加复杂性。
所以我想直接为结构体实现这个功能。
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 的功能符合我的要求。然而，我唯一关心的是它是按值传递的，因此每次都必须返回新添加的值。
是否有其他更好的方法来避免返回求和后的变量呢？
英文:
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.
答案1
得分: 33
是的，使用指针接收器：
func (a *A) AddValue(v A) {
a.value1 += v.value1
a.value2 += v.value2
}
通过使用指针接收器，将传递类型为A的值的地址，因此如果修改指向的对象，不需要返回它，将修改“原始”对象而不是副本。
你也可以简单地将其命名为Add()。为了一致性，你还可以将其参数设置为指针：
func (a *A) Add(v *A) {
a.value1 += v.value1
a.value2 += v.value2
}
然后使用它：
x, y := &A{1, 2}, &A{3, 4}
x.Add(y)
fmt.Println(x) // 输出 &{4 6}
注意
请注意，即使现在你有一个指针接收器，如果非指针值是可寻址的，仍然可以在其上调用Add()方法，因此以下示例也可以工作：
a, b := A{1, 2}, A{3, 4}
a.Add(&b)
fmt.Println(a)
a.Add()是(&a).Add()的简写形式。在Go Playground上尝试一下。
英文:
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.
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