Relationship between slice and array

A slice isn't an array, it's a struct which consists of a pointer to array element, and length of segment, and arr capacity(length). So you can consider it uses pointer and length to slice an array. When you manipulate a slice, it will return new slice, but the array under the hood may not allocate. This may cause side effect.

You can find the source code of slice here:

type slice struct {
    array unsafe.Pointer
    len   int
    cap   int
}

Let's write some code and draw a diagram to represent the concept.

s1:=[]int{1,2,3}
// cut slice : change len
s2:= s1[:2]
// cut slice : change len and pointer to array element
s3 := s1[1:3]

Potential issue -- manipulate the same array

Sometimes bug will happen when you manipulate slices both point to the same array. Like the following code, I just want to change the value in S2 only, but it affects s1 and s3, because they both point to the same array.

func main() {
    s1 := []int{1, 2, 3}    
    s2 := s1[:2]
    s3 := s1[1:3]
    fmt.Printf("%v\n%v\n%v\n", s1, s2, s3)
    fmt.Println("-----")

    s2[1] = 0
    fmt.Printf("%v\n%v\n%v\n", s1, s2, s3)
}

output:
[1 2 3]
[1 2]
[2 3]
----------
[1 0 3]
[1 0]
[0 3]

How to avoid?

• just work on the last slice. If you don't need other slice, of course it has no risk, but if you can't, there's two function to copy new slice.

• use built-in function "func copy(dst, src []Type) int",
  copy function will only copy elements, based on len of dst, from src
  

func main() {
    s1 := []int{1, 2, 3}
    s2 := make([]int,2)
    copy(s2, s1)
    s3 := make([]int,2)
    copy(s3, s1[1:])
    fmt.Printf("%v\n%v\n%v\n", s1, s2, s3)
    fmt.Println("-----")

    s2[1] = 0
    fmt.Printf("%v\n%v\n%v\n", s1, s2, s3)
}

output:
[1 2 3]
[1 2]
[2 3]
-----
[1 2 3]
[1 0]
[2 3]

• use built-in function "func append(slice []Type, elems ...Type) []Type", when you append to empty slice, it will allocate the array and make a copy.

func main() {
    s1 := []int{1, 2, 3}
    s2 := append([]int{}, s1...)
    s3 := append([]int{}, s1[1:]...)
    fmt.Printf("%v\n%v\n%v\n", s1, s2, s3)
    fmt.Println("-----")

    s2[1] = 0
    fmt.Printf("%v\n%v\n%v\n", s1, s2, s3)
}
output:
[1 2 3]
[1 2]
[2 3]
-----
[1 2 3]
[1 0]
[2 3]

Personally, I prefer to use append, because it's shorter and less code, but append can also make another issue because of allocating. Let's discuss the allocating mechanism in next article.