Golang append an item to a slice

Question

Why does the slice a remain the same? Does append() generate a new slice?

package main

import (
    "fmt"
)

var a = make([]int, 7, 8)

func Test(slice []int) {
    slice = append(slice, 100)
    fmt.Println(slice)
}

func main() {
    for i := 0; i < 7; i++ {
        a[i] = i
    }

    Test(a)
    fmt.Println(a)
}

Output:

[0 1 2 3 4 5 6 100]
[0 1 2 3 4 5 6]

Answer 1

In your example the slice argument of the Test function receives a copy of the variable a in the caller's scope.

Since a slice variable holds a "slice descriptor" which merely references an underlying array, in your Test function you modify the slice descriptor held in the slice variable several times in a row, but this does not affect the caller and its a variable.

Inside the Test function, the first append reallocates the backing array under the slice variable, copies its original contents over, appends 100 to it, and that's what you're observing. Upon exiting from Test, the slice variable goes out of scope and so does the (new) underlying array that slice references. (Jeff Lee is correct about that it's not what really happens, so the updated version follows; as he correctly states, this answer is correct, if maybe a bit too terse.)

Outside the Test function, a slice of length 7 and capacity 8 is allocated, and its 7 elements filled.
Inside the Test function, the first append sees the that the slice's capacity is still one element larger than its length — in other words, there is room for one more element to add without reallocation. So it "eats up" that remaining element and places 100 to it, after which it adjusts the length in the copy of the slice descriptor to become equal to the slice's capaticy. This does not affect the slice descriptor's in the caller's scope.

And that's what you're observing. Upon exiting from Test, the slice variable goes out of scope and so does the (new) underlying array that slice references.

If you want to make Test behave like append, you have to return the new slice from it — just like append does — and require the callers of Test to use it in the same way they would use append:

func Test(slice []int) []int {
    slice = append(slice, 100)

    fmt.Println(slice)

    return slice
}

a = Test(a)

Please read this article thoroughly as it basically shows you how to implement append by hand, after explaining how slices are working internally. Then read this.

Answer 2

Typical append usage is

a = append(a, x)

because append may either modify its argument in-place or return a copy of its argument with an additional entry, depending on the size and capacity of its input. Using a slice that was previously appended to may give unexpected results, e.g.

a := []int{1,2,3}
a = append(a, 4)
fmt.Println(a)
append(a[:3], 5)
fmt.Println(a)

may print

[1 2 3 4]
[1 2 3 5]

Answer 3

In order to make your code work without having to return the slice from Test, you can pass a pointer like this:

package main

import (
    "fmt"
)

var a = make([]int, 7, 8)

func Test(slice *[]int) {
    *slice = append(*slice, 100)

    fmt.Println(*slice)
}

func main() {

    for i := 0; i < 7; i++ {
        a[i] = i
    }

    Test(&a)

    fmt.Println(a)
}

Answer 4

NOTICE that append generates a new slice if cap is not sufficient. @kostix's answer is correct, or you can pass slice argument by pointer!

Answer 5

Try this, which I think makes it clear. the underlying array is changed but our slice is not, print just prints len() chars, by another slice to the cap(), you can see the changed array:

func main() {

  for i := 0; i < 7; i++ {
      a[i] = i
  }

  Test(a)

  fmt.Println(a) // prints [0..6]
  fmt.Println(a[:cap(a)] // prints [0..6,100]
}

Answer 6

Explanation (read inline comments):

package main

import (
    "fmt"
)

var a = make([]int, 7, 8)
// A slice is a descriptor of an array segment. 
// It consists of a pointer to the array, the length of the segment, and its capacity (the maximum length of the segment).
// The length is the number of elements referred to by the slice.
// The capacity is the number of elements in the underlying array (beginning at the element referred to by the slice pointer).
// |-> Refer to: https://blog.golang.org/go-slices-usage-and-internals -> "Slice internals" section

func Test(slice []int) {
    // slice receives a copy of slice `a` which point to the same array as slice `a`
    slice[6] = 10
    slice = append(slice, 100)
    // since `slice` capacity is 8 & length is 7, it can add 100 and make the length 8
    fmt.Println(slice, len(slice), cap(slice), " << Test 1")
    slice = append(slice, 200)
    // since `slice` capacity is 8 & length also 8, slice has to make a new slice 
    // - with double of size with point to new array (see Reference 1 below).
    // (I'm also confused, why not (n+1)*2=20). But make a new slice of 16 capacity).
    slice[6] = 13 // make sure, it's a new slice :)
    fmt.Println(slice, len(slice), cap(slice), " << Test 2")
}

func main() {
    for i := 0; i < 7; i++ {
        a[i] = i
    }

    fmt.Println(a, len(a), cap(a))
    Test(a)
    fmt.Println(a, len(a), cap(a))
    fmt.Println(a[:cap(a)], len(a), cap(a))
    // fmt.Println(a[:cap(a)+1], len(a), cap(a)) -> this'll not work
}

Output:

[0 1 2 3 4 5 6] 7 8
[0 1 2 3 4 5 10 100] 8 8  << Test 1
[0 1 2 3 4 5 13 100 200] 9 16  << Test 2
[0 1 2 3 4 5 10] 7 8
[0 1 2 3 4 5 10 100] 7 8

Reference 1: https://blog.golang.org/go-slices-usage-and-internals

func AppendByte(slice []byte, data ...byte) []byte {
    m := len(slice)
    n := m + len(data)
    if n > cap(slice) { // if necessary, reallocate
        // allocate double what's needed, for future growth.
        newSlice := make([]byte, (n+1)*2)
        copy(newSlice, slice)
        slice = newSlice
    }
    slice = slice[0:n]
    copy(slice[m:n], data)
    return slice
}

Answer 7

Go takes a more lean and lazy approach in doing this. It keeps modifying the same underlying array until the capacity of a slice is reached.

Ref: http://criticalindirection.com/2016/02/17/slice-with-a-pinch-of-salt/

Output of the example from the link explains the behavior of slices in Go.

Creating slice a.

Slice a len=7 cap=7 [0 0 0 0 0 0 0]

Slice b refers to the 2, 3, 4 indices in slice a. Hence, the capacity is 5 (= 7-2).

b := a[2:5]
Slice b len=3 cap=5 [0 0 0]

Modifying slice b, also modifies a, since they are pointing to the same underlying array.

b[0] = 9
Slice a len=7 cap=7 [0 0 9 0 0 0 0]
Slice b len=3 cap=5 [9 0 0]

Appending 1 to slice b. Overwrites a.

Slice a len=7 cap=7 [0 0 9 0 0 1 0]
Slice b len=4 cap=5 [9 0 0 1]

Appending 2 to slice b. Overwrites a.

Slice a len=7 cap=7 [0 0 9 0 0 1 2]
Slice b len=5 cap=5 [9 0 0 1 2]

Appending 3 to slice b. Here, a new copy is made as the capacity is overloaded.

Slice a len=7 cap=7 [0 0 9 0 0 1 2]
Slice b len=6 cap=12 [9 0 0 1 2 3]

Verifying slices a and b point to different underlying arrays after the capacity-overload in the previous step.

b[1] = 8
Slice a len=7 cap=7 [0 0 9 0 0 1 2]
Slice b len=6 cap=12 [9 8 0 1 2 3]

Answer 8

package main

import (
    "fmt"
)

func a() {
    x := []int{}
    x = append(x, 0)
    x = append(x, 1)  // commonTags := labelsToTags(app.Labels)
    y := append(x, 2) // Tags: append(commonTags, labelsToTags(d.Labels)...)
    z := append(x, 3) // Tags: append(commonTags, labelsToTags(d.Labels)...)
    fmt.Println(y, z)
}

func b() {
    x := []int{}
    x = append(x, 0)
    x = append(x, 1)
    x = append(x, 2)  // commonTags := labelsToTags(app.Labels)
    y := append(x, 3) // Tags: append(commonTags, labelsToTags(d.Labels)...)
    z := append(x, 4) // Tags: append(commonTags, labelsToTags(d.Labels)...)
    fmt.Println(y, z)
}

func main() {
    a()
    b()
}

First guess could be

[0, 1, 2] [0, 1, 3]
[0, 1, 2, 3] [0, 1, 2, 4]

but in fact it results in

[0, 1, 2] [0, 1, 3]
[0, 1, 2, 4] [0, 1, 2, 4]

More details see https://allegro.tech/2017/07/golang-slices-gotcha.html

Answer 9

Answer: append() will return new underlying array if has no sufficient capacity. In your example,

var a = make([]int, 7, 8)

You allocate a slice (length is 7)of an underlying array(capacity is 8) to a, then pass it to function as parameter slice. When append() is called, it found the 1 capacity, then just update the slice's len from 7 to 8 and put the value 100 into that position.

The slice a is different to the slice slice, having different len property. len and cap are the property of slice, not the underlying array. For more details: Are slices passed by value?.

Run the example below:

package main

import (
    "fmt"
)

var a = make([]int, 7, 8)

func Test(slice []int) {
    fmt.Printf("slice's address is %p\n", &slice)
    fmt.Println("slice: cap=",cap(slice),"len=",len(slice))
    slice = append(slice, 100)
    fmt.Println("slice: cap=",cap(slice),"len=",len(slice))
    fmt.Println(slice)
}

func main() {
    for i := 0; i < 7; i++ {
        a[i] = i
    }
    fmt.Printf("a's address is %p\n", &a)
    fmt.Println("a: cap=",cap(a),"len=",len(a))
    Test(a)
    fmt.Println("a: cap=",cap(a),"len=",len(a))
    fmt.Println(a)
    fmt.Println(a[:8]) // manully extend a's len to cap of 8
}

Result is:

❯❯  Temp  17:33  go run .\test.go
a's address is 0x2cbfc0
a: cap= 8 len= 7
slice's address is 0xc000098060
slice: cap= 8 len= 7
slice: cap= 8 len= 8
[0 1 2 3 4 5 6 100]
a: cap= 8 len= 7
[0 1 2 3 4 5 6]
[0 1 2 3 4 5 6 100]

Answer 10

I think the original answer is not exactly correct. append() changed both the slices and the underlying array even though the underlying array is changed but still shared by both of the slices.

As specified by the Go Doc:

A slice does not store any data, it just describes a section of an underlying array. (Link)

Slices are just wrapper values around arrays, meaning that they contain information about how they slice an underlying array which they use to store a set of data. Therefore, by default, a slice, when passed to another method, is actually passed by value, instead of reference/pointer even though they will still be using the same underlying array. Normally, arrays are also passed by value too, so I assume a slice points at an underlying array instead of store it as a value. Regarding your question, when you run passed your slice to the following function:

func Test(slice []int) {
    slice = append(slice, 100)
    fmt.Println(slice)
}

you actually passed a copy of your slice along with a pointer to the same underlying array.That means, the changes you did to the slice didn't affect the one in the main function. It is the slice itself which stores the information regarding how much of an array it slices and exposes to the public. Therefore, when you ran append(slice, 1000), while expanding the underlying array, you also changed slicing information of slice too, which was kept private in your Test() function.

However, if you have changed your code as follows, it might have worked:

func main() {
    for i := 0; i < 7; i++ {
        a[i] = i
    }

    Test(a)
    fmt.Println(a[:cap(a)])
}

The reason is that you expanded a by saying a[:cap(a)] over its changed underlying array, changed by Test() function. As specified here:

You can extend a slice's length by re-slicing it, provided it has sufficient capacity. (Link)

Answer 11

Yes, when you use the append(), to add values to a Slice, it usually creates a new Slice and doesn't overwrite the original Slice.

Check the below code snippet

package main

import "fmt"

func main() {

    var ages = []int{3,6,8,1,9}

    fmt.Println(append(ages, 13))

    fmt.Println("Original Slice is still: ", ages)
}

If you need to overwrite the original Slice, you need to set the append() to the Slice name as below.

ages = append(ages, 12)
fmt.Println("Now original Slice is: ", ages)

Answer 12

Very simple.

1. If the capacity of the slice is enough to append, then just use the existing underlying array.

2. If the capacity of the slice is not enough to append, then create a new underlying array.

Answer 13

Here is a nice implementation of append for slices. I guess its similar to what is going on under the hood:

package main

import "fmt"

func main() {
    slice1 := []int{0, 1, 2, 3, 4}
    slice2 := []int{55, 66, 77}
    fmt.Println(slice1)
    slice1 = Append(slice1, slice2...) // The '...' is essential!
    fmt.Println(slice1)
}

// Append ...
func Append(slice []int, items ...int) []int {
    for _, item := range items {
        slice = Extend(slice, item)
    }
    return slice
}

// Extend ...
func Extend(slice []int, element int) []int {
    n := len(slice)
    if n == cap(slice) {
        // Slice is full; must grow.
        // We double its size and add 1, so if the size is zero we still grow.
        newSlice := make([]int, len(slice), 2*len(slice)+1)
        copy(newSlice, slice)
        slice = newSlice
    }
    slice = slice[0 : n+1]
    slice[n] = element
    return slice
}

Answer 14

Append to end of a slice, or create a new entry if slice is empty

// in := [][]int{{}}, in := [][]int{{1,3},{2,3}}
// addtoEndofSliceArray(in,10)
// out=[[10]], out=[[1,3],[2,3,10]]

func addtoEndofSliceArray(in [][]int,element int)(out [][]int){
    if len(in) >0 {
        k :=in[len(in)-1]
        k = append(k,element)
        in = in[:len(in)-1]
        in = append(in, k)
    }else{
        in = [][]int{{element}}
    }
    return in
}