为什么在为golang类型实现sort.Interface时不需要方法接收者是指针？

Question

I am reading the docs for the sort stdlib package and the sample code reads like this:

type ByAge []Person

func (a ByAge) Len() int           { return len(a) }
func (a ByAge) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a ByAge) Less(i, j int) bool { return a[i].Age < a[j].Age }

As I've learnt, function that mutate a type T needs to use *T as its method receiver. In the case of Len, Swap and Less why does it work ? Or am I misunderstanding the difference between using T vs *T as method receivers ?

Answer 1

Go has three reference types:

• map
• slice
• channel

Every instance of these types holds a pointer to the actual data internally. This means that when you pass a value of one of these types the value is copied like every other value but the internal pointer still points to the same value.

Quick example (run on play):

func dumpFirst(s []int) {
    fmt.Printf("address of slice var: %p, address of element: %p
", &s, &s[0])
}

s1 := []int{1, 2, 3}
s2 := s1

dumpFirst(s1)
dumpFirst(s2)

will print something like:

address of slice var: 0x1052e110, address of element: 0x1052e100
address of slice var: 0x1052e120, address of element: 0x1052e100

You can see: the address of the slice variable changes but the address of the first element in that slice remains the same.