(Go) do not use append to insert elements

This is a creation in Article, where the information may have evolved or changed.

The title seems a little bluffing ....
The cause was that I was doing an insert function for slice, and I did some performance testing for several versions of the package. The results are unexpected!

Import ("FMT"    "Reflect"    "Time") Func Copyinsert (Slice interface{}, pos int, Value interface{}) interface{} {V: = Reflect. ValueOf(slice) v = reflect. Append(V, reflect. ValueOf(value)) Reflect. Copy(V. Slice(pos+1V. Len()), V. Slice(POS, V. Len())) v. Index(POS). Set(Reflect. ValueOf(value)) Return V. Interface()}func Insert (Slice interface{}, pos int, Value interface{}) interface{} {V: = Reflect. ValueOf(slice) NE: = reflect. Makeslice(Reflect. Sliceof(Reflect. TypeOf(value)),1,1) NE. Index(0). Set(Reflect. ValueOf(value)) v = Reflect. Appendslice(V. Slice(0, POS), reflect. Appendslice(NE, V. Slice(POS, V. Len())) return V. Interface()}func main () {slice: = []int{1,2} slice2: = []int{1,2} slice3: = []int{1,2} T0: = Time. now() for I: =1; i < 10000; i++ {Slice = Append (slice[:1], append ([]int{i}, slice[1:]...)    ...) } T1: = time. now() for I: =1; i < 10000; i++ {Slice2 = Insert (Slice2,1, i).    ([]int)} T2: = Time. now() for I: =1; i < 10000; i++ {Slice3 = Copyinsert (Slice3,1, i). ([]int)//FMT. Println(SLICE3)} T3: = time. now()//element detection for I: =0; i < 10000; i++ {If slice[i]! = Slice2[i] | | Slice2[i]! = Slice3[i] {FMT. Println("Error")}} FMT. Println("reflect append insert:", T2. Sub(t1),"Append insert:", T1. Sub(t0),"Copy Insert:", T3. Sub(T2))}

Output results

append insert: 273.276271append 195.938231copy 29.626058ms

The use of reflective reflect than ordinary append is understandable, but actually 10 times times slower than copy Instert. >_<.

I didn't believe it, so I did the test.

Import ("FMT"    "Time") Func Main () {s1: = []int{1,2,3} s2: = []int{1,2,3}//Insert the third T1: = time. Now () forI: =0; I <10000; i++ {S1 = append (s1[0:3], append ([]int{i}, s1[3:]...)...)} t2: = time. Now ()//Every time you insert a third forI: =0; I <10000; i++ {s2 = append (s2, i) copy (s2[3: Len (S2)], s2[2: Len (S2)-1]) s2[3] = i} t3: = time. Now ()//Detection error forI: =0; I < Len (S1); i++ {ifS1[i]! = S2[i] {fmt. Println ("Error")}} FMT. Println (T3. Sub (T2), T2. Sub (t1))}

Output results

copy instert: 21.245765append 189.930674ms

The result is still 10 times times worse!

Guess: two times append (), doing slice slices will produce the underlying array copy. (Append ([]int{i}, S1[3:] ...) ), but when the S1 is very large, the more copying, the worse the performance, and copy is using the original underlying array. >_<.

Welcome to Daniel's advice ....