append - Concatenate two slices in Go

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 := [10]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 := [10]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.