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 source slice into a destination slice. (As a special case, it also will copy bytes from a string to a slice of bytes.) The source and destination may overlap. Copy returns the number of elements copied, which will be the minimum of len(src) and len(dst).
But when I do:
arr := []int{1, 2, 3}
tmp := []int{}
copy(tmp, arr)
fmt.Println(tmp)
fmt.Println(arr)
My tmp
is empty as it was before (I even tried to use arr, tmp
):
[]
[1 2 3]
You can check it on go playground. So why can not I copy a slice?
dst
should be at least as big as many elements you want to copy (for a full copy of src
it means len(dst) >= len(src)
).
b := append([]int{}, a...)
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).
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]
As pointed out in the comments below, append
may allocate excess memory if the slice isn't sized correctly to begin with. A nice solution to this is to preallocate a slice of the right capacity, like so:
tmp := append(make([]int, 0, len(arr)), arr...)
So a shorthand for copying array arr
would be append(make([]int, 0, len(arr)), arr...)
https://play.golang.org/p/xwevI1chGrd
tmp := append(make([]int, 0, len(arr)), arr...)
. Under the hood, the append
function checks the capacity of the slice on each call, allocating a new slice if the capacity is about to be exceeded. This solution has the added benefit of circumventing the repeated re-allocations.
copy()
from my benchmarks is about 15%
faster. OTOH copy()
is more complicated to use, as this Go Playground illustrates go.dev/play/p/XDGuVU0xfRy — Benchmarks and results here gist.github.com/mikeschinkel/fbeb291b90970581b4428fbc14e4ab19 as you can can't benchmark reliably on a shared server. HOWEVER, I would normally still use the append()
method because it is easier to understand and less likely to break than the copy()
version unless performance is absolutely critical for a given use-case.
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
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
The best way to clone as slice is
sClone = append(s[:0:0], s...)
This implementation has two advantage:
make sure that the result sClone is nil if s is nil, and is not nil if s is not nil. No need to import the containing package of type T even if T is declared in another package
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]
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]
Sweet, Simple, Performant, No need to be careful of length, No Memory overlap, Different copies
slice2 := append([]int{}, slice1...)
If you don't care about the speed:
import "golang.org/x/exp/slices"
tmp := slices.Clone(arr)
With Go 1.18 and generics, any slices now could be copied with slices.Clone
from package "golang.org/x/exp/slices"
. Playground
Success story sharing
copy
doesn't contain logic for growing the destination slice if the destination slice is too small, but there is another built-in function that does:append
While in this example it's better just to allocate the right sized slice in the first place,append
can be used when you already have a slice and want to grow it by adding elements to the end.