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## Concatenate two slices in Go

### Question

I'm trying to combine the slice `[1, 2]` and the slice `[3, 4]`. How can I do this in Go?

I tried:

``````append([]int{1,2}, []int{3,4})
``````

but got:

``````cannot use []int literal (type []int) as type int in append
``````

However, the documentation seems to indicate this is possible, what am I missing?

``````slice = append(slice, anotherSlice...)
``````
2016/10/14
1
500
10/14/2016 7:05:29 AM

### Accepted Answer

Add dots after the second slice:

``````//---------------------------vvv
append([]int{1,2}, []int{3,4}...)
``````

This is just like any other variadic function.

``````func foo(is ...int) {
for i := 0; i < len(is); i++ {
fmt.Println(is[i])
}
}

func main() {
foo([]int{9,8,7,6,5}...)
}
``````
2015/03/05
921
3/5/2015 7:40:00 PM

Appending to and copying slices

The variadic function `append` appends zero or more values `x` to `s` of type `S`, which must be a slice type, and returns the resulting slice, also of type `S`. The values `x` are passed to a parameter of type `...T` where `T` is the element type of `S` and the respective parameter passing rules apply. As a special case, append also accepts a first argument assignable to type `[]byte` with a second argument of `string` type followed by `...`. This form appends the bytes of the string.

``````append(s S, x ...T) S  // T is the element type of S

s0 := []int{0, 0}
s1 := append(s0, 2)        // append a single element     s1 == []int{0, 0, 2}
s2 := append(s1, 3, 5, 7)  // append multiple elements    s2 == []int{0, 0, 2, 3, 5, 7}
s3 := append(s2, s0...)    // append a slice              s3 == []int{0, 0, 2, 3, 5, 7, 0, 0}
``````

Passing arguments to ... parameters

If `f` is variadic with final parameter type `...T`, then within the function the argument is equivalent to a parameter of type `[]T`. At each call of `f`, the argument passed to the final parameter is a new slice of type `[]T` whose successive elements are the actual arguments, which all must be assignable to the type `T`. The length of the slice is therefore the number of arguments bound to the final parameter and may differ for each call site.

The answer to your question is example `s3 := append(s2, s0...)` in the Go Programming Language Specification. For example,

``````s := append([]int{1, 2}, []int{3, 4}...)
``````
2013/04/27

Nothing against the other answers, but I found the brief explanation in the docs more easily understandable than the examples in them:

## func append

`func append(slice []Type, elems ...Type) []Type` The append built-in function appends elements to the end of a slice. If it has sufficient capacity, the destination is resliced to accommodate the new elements. If it does not, a new underlying array will be allocated. Append returns the updated slice. It is therefore necessary to store the result of append, often in the variable holding the slice itself:

``````slice = append(slice, elem1, elem2)
slice = append(slice, anotherSlice...)
``````

As a special case, it is legal to append a string to a byte slice, like this:

``````slice = append([]byte("hello "), "world"...)
``````
2015/04/17

I think it's important to point out and to know that if the destination slice (the slice you append to) has sufficient capacity, the append will happen "in-place", by reslicing the destination (reslicing to increase its length in order to be able to accommodate the appendable elements).

This means that if the destination was created by slicing a bigger array or slice which has additional elements beyond the length of the resulting slice, they may get overwritten.

To demonstrate, see this example:

``````a := int{1, 2}
fmt.Printf("a: %v\n", a)

x, y := a[:2], []int{3, 4}
fmt.Printf("x: %v, y: %v\n", x, y)
fmt.Printf("cap(x): %v\n", cap(x))

x = append(x, y...)
fmt.Printf("x: %v\n", x)

fmt.Printf("a: %v\n", a)
``````

Output (try it on the Go Playground):

``````a: [1 2 0 0 0 0 0 0 0 0]
x: [1 2], y: [3 4]
cap(x): 10
x: [1 2 3 4]
a: [1 2 3 4 0 0 0 0 0 0]
``````

We created a "backing" array `a` with length `10`. Then we create the `x` destination slice by slicing this `a` array, `y` slice is created using the composite literal `[]int{3, 4}`. Now when we append `y` to `x`, the result is the expected `[1 2 3 4]`, but what may be surprising is that the backing array `a` also changed, because capacity of `x` is `10` which is sufficient to append `y` to it, so `x` is resliced which will also use the same `a` backing array, and `append()` will copy elements of `y` into there.

If you want to avoid this, you may use a full slice expression which has the form

``````a[low : high : max]
``````

which constructs a slice and also controls the resulting slice's capacity by setting it to `max - low`.

See the modified example (the only difference is that we create `x` like this: `x = a[:2:2]`:

``````a := int{1, 2}
fmt.Printf("a: %v\n", a)

x, y := a[:2:2], []int{3, 4}
fmt.Printf("x: %v, y: %v\n", x, y)
fmt.Printf("cap(x): %v\n", cap(x))

x = append(x, y...)
fmt.Printf("x: %v\n", x)

fmt.Printf("a: %v\n", a)
``````

Output (try it on the Go Playground)

``````a: [1 2 0 0 0 0 0 0 0 0]
x: [1 2], y: [3 4]
cap(x): 2
x: [1 2 3 4]
a: [1 2 0 0 0 0 0 0 0 0]
``````

As you can see, we get the same `x` result but the backing array `a` did not change, because capacity of `x` was "only" `2` (thanks to the full slice expression `a[:2:2]`). So to do the append, a new backing array is allocated that can store the elements of both `x` and `y`, which is distinct from `a`.

2018/01/16

I would like to emphasize @icza answer and simplify it a bit since it is a crucial concept. I assume that reader is familiar with slices.

``````c := append(a, b...)
``````

This is a valid answer to the question. BUT if you need to use slices 'a' and 'c' later in code in different context, this is not the safe way to concatenate slices.

To explain, lets read the expression not in terms of slices, but in terms of underlying arrays:

"Take (underlying) array of 'a' and append elements from array 'b' to it. If array 'a' has enough capacity to include all elements from 'b' - underlying array of 'c' will not be a new array, it will actually be array 'a'. Basically, slice 'a' will show len(a) elements of underlying array 'a', and slice 'c' will show len(c) of array 'a'."

append() does not necessarily create a new array! This can lead to unexpected results. See Go Playground example.

Always use make() function if you want to make sure that new array is allocated for the slice. For example here are few ugly but efficient enough options for the task.

``````la := len(a)
c := make([]int, la, la + len(b))
_ = copy(c, a)
c = append(c, b...)
``````

``````la := len(a)
c := make([]int, la + len(b))
_ = copy(c, a)
_ = copy(c[la:], b)
``````
2019/11/07

## append( ) function and spread operator

Two slices can be concatenated using `append` method in the standard golang library. Which is similar to the `variadic` function operation. So we need to use `...`

``````package main

import (
"fmt"
)

func main() {
x := []int{1, 2, 3}
y := []int{4, 5, 6}
z := append([]int{}, append(x, y...)...)
fmt.Println(z)
}
``````

output of the above code is: [1 2 3 4 5 6]

2018/11/07

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