英文:
Are these two Go pieces of code equivalent?
问题
这两个函数是等价的，它们都用于设置 Square 结构体的 Side 字段。没有内部差异或其他特殊之处。
英文:
Having this struct
type Square struct {
Side int
}
Are these to functions equivalent?
func (s *Square) SetSide(side int) {
s.Side = side
}
vs
func SetSquareSide(s *Square, side int) {
s.Side = side
}
I know they do the same, but are they really equivalent? I mean, is there any internal difference or something?
Try online: https://play.golang.org/p/gpt2KmsVrz
答案1
得分: 6
据我所知，它们的工作方式是相同的。
其中一个区别是只有第一个能够满足接口规范。
英文:
As far as I know they work the same way.
One difference is that only the first one could satisfy an interface specification.
答案2
得分: 3
这些"函数"的工作方式相同，在实际中它们以几乎相同的方式被调用。该方法被称为方法表达式，接收器作为第一个参数：
var s Square
// 方法调用
s.SetSide(5)
// 等同于方法表达式
(*Square).SetSide(&s, 5)
SetSide 方法也可以作为方法值用于满足函数签名 func(int)，而 SetSquareSide 则不能。
var f func(int)
f = a.SetSide
f(9)
除此之外，显而易见的是 Square 的方法集满足接口
interface {
SetSide(int)
}
英文:
These "function" the same way, and in reality they are called in nearly the same way. The method is called as a method expression, with the receiver as the first argument:
var s Square
// The method call
s.SetSide(5)
// is equivalent to the method expression
(*Square).SetSide(&s, 5)
The SetSide method can also be used as a method value to satisfy the function signature func(int), where as the SetSquareSide cannot.
var f func(int)
f = a.SetSide
f(9)
This is on top of the obvious fact that the method set of Square satisfies the interface
interface {
SetSide(int)
}
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