通过上面的代码, AddValue 按我的意愿工作。然而,我唯一担心的是它是一种价值传递,因此我必须每次都返回新增值。
是否还有其他更好的方法,以便避免返回总和变量。
解决方案是的,使用指针接收器:
func(a * A)AddValue(v A){
a.value1 + = v.value1
a.value2 + = v.value2
$ b
通过使用指针接收器,类型为
会被传递,因此如果你修改了指向的对象,你不必返回它,你将修改原始对象而不是副本。A
您也可以简单地将它命名为
Add()
。 (b)
$ b $ pre $ func(a * A)Add(v * A) {
a.value1 + = v.value1
a.value2 + = v.value2
}
使用它:
x,y:=& A {1,2 },& A {3,4}
x.Add(y)
fmt.Println(x)//打印& {4 6}
注意
Add()
a,b:= A {1,2},A {3,4}
a.Add & b)
fmt.Println(a)
a .Add()(& a).Add()
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)
}
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.
Yes, use pointer receiver:
func (a *A) AddValue(v A) {
a.value1 += v.value1
a.value2 += v.value2
}
A
Add()
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
Add()
a, b := A{1, 2}, A{3, 4}
a.Add(&b)
fmt.Println(a)
a.Add()(&a).Add()
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