Why can't I duplicate a slice with `copy()`?

Question

I need to make a copy of a slice in Go and reading the docs there is a copy function at my disposal.

The copy built-in function copies elements from a so

Answer 1

NOTE: This is an incorrect solution as @benlemasurier proved

Here is a way to copy a slice. I'm a bit late, but there is a simpler, and faster answer than @Dave's. This are the instructions generated from a code like @Dave's. These is the instructions generated by mine. As you can see there are far fewer instructions. What is does is it just does append(slice), which copies the slice. This code:

package main

import "fmt"

func main() {
    var foo = []int{1, 2, 3, 4, 5}
    fmt.Println("foo:", foo)
    var bar = append(foo)
    fmt.Println("bar:", bar)
    bar = append(bar, 6)
    fmt.Println("foo after:", foo)
    fmt.Println("bar after:", bar)
}

Outputs this:

foo: [1 2 3 4 5]
bar: [1 2 3 4 5]
foo after: [1 2 3 4 5]
bar after: [1 2 3 4 5 6]

Answer 2

The copy() runs for the least length of dst and src, so you must initialize the dst to the desired length.

A := []int{1, 2, 3}
B := make([]int, 3)
copy(B, A)
C := make([]int, 2)
copy(C, A)
fmt.Println(A, B, C)

Output:

[1 2 3] [1 2 3] [1 2]

You can initialize and copy all elements in one line using append() to a nil slice.

x := append([]T{}, []...)

Example:

A := []int{1, 2, 3}
B := append([]int{}, A...)
C := append([]int{}, A[:2]...)
fmt.Println(A, B, C)    

Output:

[1 2 3] [1 2 3] [1 2]

Comparing with allocation+copy(), for greater than 1,000 elements, use append. Actually bellow 1,000 the difference may be neglected, make it a go for rule of thumb unless you have many slices.

BenchmarkCopy1-4                50000000            27.0 ns/op
BenchmarkCopy10-4               30000000            53.3 ns/op
BenchmarkCopy100-4              10000000           229 ns/op
BenchmarkCopy1000-4              1000000          1942 ns/op
BenchmarkCopy10000-4              100000         18009 ns/op
BenchmarkCopy100000-4              10000        220113 ns/op
BenchmarkCopy1000000-4              1000       2028157 ns/op
BenchmarkCopy10000000-4              100      15323924 ns/op
BenchmarkCopy100000000-4               1    1200488116 ns/op
BenchmarkAppend1-4              50000000            34.2 ns/op
BenchmarkAppend10-4             20000000            60.0 ns/op
BenchmarkAppend100-4             5000000           240 ns/op
BenchmarkAppend1000-4            1000000          1832 ns/op
BenchmarkAppend10000-4            100000         13378 ns/op
BenchmarkAppend100000-4            10000        142397 ns/op
BenchmarkAppend1000000-4            2000       1053891 ns/op
BenchmarkAppend10000000-4            200       9500541 ns/op
BenchmarkAppend100000000-4            20     176361861 ns/op

Answer 3

Another simple way to do this is by using append which will allocate the slice in the process.

arr := []int{1, 2, 3}
tmp := append([]int(nil), arr...)  // Notice the ... splat
fmt.Println(tmp)
fmt.Println(arr)

Output (as expected):

[1 2 3]
[1 2 3]

So a shorthand for copying array arr would be append([]int(nil), arr...)

https://play.golang.org/p/sr_4ofs5GW

Answer 4

If your slices were of the same size, it would work:

arr := []int{1, 2, 3}
tmp := []int{0, 0, 0}
i := copy(tmp, arr)
fmt.Println(i)
fmt.Println(tmp)
fmt.Println(arr)

Would give:

3
[1 2 3]
[1 2 3]

From "Go Slices: usage and internals":

The copy function supports copying between slices of different lengths (it will copy only up to the smaller number of elements)

The usual example is:

t := make([]byte, len(s), (cap(s)+1)*2)
copy(t, s)
s = t

Answer 5

The Go Programming Language Specification

Appending to and copying slices

The function copy copies slice elements from a source src to a destination dst and returns the number of elements copied. Both arguments must have identical element type T and must be assignable to a slice of type []T. The number of elements copied is the minimum of len(src) and len(dst). As a special case, copy also accepts a destination argument assignable to type []byte with a source argument of a string type. This form copies the bytes from the string into the byte slice.

copy(dst, src []T) int
copy(dst []byte, src string) int

tmp needs enough room for arr. For example,

package main

import "fmt"

func main() {
    arr := []int{1, 2, 3}
    tmp := make([]int, len(arr))
    copy(tmp, arr)
    fmt.Println(tmp)
    fmt.Println(arr)
}

Output:

[1 2 3]
[1 2 3]

Answer 6

The builtin copy(dst, src) copies min(len(dst), len(src)) elements.

So if your dst is empty (len(dst) == 0), nothing will be copied.

Try tmp := make([]int, len(arr)) (Go Playground):

arr := []int{1, 2, 3}
tmp := make([]int, len(arr))
copy(tmp, arr)
fmt.Println(tmp)
fmt.Println(arr)

Output (as expected):

[1 2 3]
[1 2 3]

Unfortunately this is not documented in the builtin package, but it is documented in the Go Language Specification: Appending to and copying slices:

The number of elements copied is the minimum of len(src) and len(dst).

Edit:

Finally the documentation of copy() has been updated and it now contains the fact that the minimum length of source and destination will be copied:

Copy returns the number of elements copied, which will be the minimum of len(src) and len(dst).