golang sort slice ascending or descending

Question

I need to sort a slice of a type that is coming from a 3rdparty package. Based on some condition the order must be ascending or descending.

The solution I come up with is:

type fooAscending []foo  func (v fooAscending) Len() int           { return len(v) } func (v fooAscending) Swap(i, j int)      { v[i], v[j] = v[j], v[i] } func (v fooAscending) Less(i, j int) bool { return v[i].Amount < v[j].Amount }  type fooDescending []foo  func (v fooDescending) Len() int           { return len(v) } func (v fooDescending) Swap(i, j int)      { v[i], v[j] = v[j], v[i] } func (v fooDescending) Less(i, j int) bool { return v[i].Amount > v[j].Amount }  if someCondition {     sort.Sort(fooAscending(array)) } else {     sort.Sort(fooDescending(array)) } 

Is there a better way to do this. 13 lines of code for this task and most of it is duplicated, seems a bit too much.

Answer 1

As of Go 1.8, there is an easier way to sort a slice that does not require you to define new types. You simply pass an anonymous function to the sort.Slice function.

a := []int{5, 3, 4, 7, 8, 9} sort.Slice(a, func(i, j int) bool {     return a[i] < a[j] }) for _, v := range a {     fmt.Println(v) } 

This will sort in ascending order, if you want the opposite, simply write a[i] > a[j] in the anonymous function.

Answer 2

You're looking for sort.Reverse. That will let you say:

sort.Sort(sort.Reverse(fooAscending(s)))