Go中的Java Arrays.copyOfRange等效于什么？

Question

I'm working on the creating a Java library that deals with management of Arrays. Is there an equivalent of Arrays.copyOfRange in Go?

Answer 1

A simple one-liner would be (index check omitted):

func copyOfRange(src []byte, from, to int) []byte {
    return append([]byte(nil), src[from:to]...)
}

A simple slice expression "almost" does the job, but since Java's Arrays.copyOfRange() returns a copy independent from the source, we need to copy the slicing result to a new slice (because the result of slicing will share the backing array).

We can do that by allocating one with make(), and use the builtin copy(), or simply use append() to append it to an empty or nil slice, which will take care of the allocation and copying.

Example using the above function:

src := []byte{0, 1, 2, 3, 4, 5}
dst := copyOfRange(src, 2, 4)
fmt.Println(dst)

Output (try it on the Go Playground):

[2 3]

For completeness, this is how it would look like with make() and copy():

func copyOfRange2(src []byte, from, to int) []byte {
    src = src[from:to]
    dst := make([]byte, len(src))
    copy(dst, src)
    return dst
}

One thing to note here: the builtin append() allocates more space than needed, thinking of future growth. So if you don't plan to "grow" the returned slice, copyOfRange2() is a better option.

See this comparison:

dst := copyOfRange(src, 2, 4)
fmt.Println(dst, cap(dst))

dst = copyOfRange2(src, 2, 4)
fmt.Println(dst, cap(dst))

Output (try it on the Go Playground):

[2 3] 8
[2 3] 2

As you can see, append() (inside copyOfRange()) allocated a backing array with a size of 8, while in our copyOfRange2() we explicitly allocated a slice (and backing array) of size 2.