append（）如何在导致这种情况的幕后工作？

Question

I'm pretty new to Go and trying to track down a bug in my codebase. In the process I've distilled the problem down to a single call to append(), but can't figure out why it's behaving the way it is...

func main() {
    foo := []string{"a", "b", "c"}
    fmt.Printf("before: %v
", foo)

    i := 0
    noop(append(foo[:i], foo[i+1:]...)) // -- call append, but do nothing with the result
    fmt.Printf(" after: %v
", foo)
}

func noop(a interface{}) {} // -- avoid "evaluated but not used" errors

Try it here

So, what the heck is really going on here?

Answer 1

append(foo[:i], foo[i+1:]...) does the following:

1. It takes the foo[:i] slice, which is foo[:0] and basically a slice with length 0 and capacity (at least) 3.

2. As soon as the capacity is enough to append the values - the underlying array is reused

3. You write b and c into the indexes 0 and 1 correspondingly of the underlying array.

Then you check the foo variable that uses the underlying array we just modified and that contains the b c c values.

Compare with the following:

noop(append(foo[:i], "a", "a", "a", "a", "a"))

Here the list of values to append is longer than the current capacity. So the runtime allocates a new underlying array. And you don't mutate the foo. https://play.golang.org/p/RooYG_p9Z8