Why does a slice expression of an empty slice have values in it?

Question

Just started leaning Golang. Facing some trouble understanding the slice expression.

package main

import "fmt"

func main() {
    x := make([]int, 0, 5)
    y := append(x, 20)
    fmt.Println(x, y, x[:2], y[:2])
}

Output:

[] [20] [20 0] [20 0]

Here's my understanding of what this represents:

• x := make([]int, 0, 5) - make a new slice of ints with 0 length and 5 capacity.

• y := append(x, 20) - create a a new slice using append function with length 1 & capacity of 5. y is [20].

• fmt.Println(x, y, x[:2], y[:2])

  • How does x[:2] print [20 0]? Shouldn't it just cause an error? If this was the slice expression creating a new slice shouldn't it be [0 0]? How is the first value 20?

I am missing some fundamental concept here.

If I were to take a guess I think creating y using y := append(x, 20) is somehow sharing the underlying memory. This is not shown when I print x directly, but when I am printing x[:2] it is shown.

Answer 1

y := append(x, 20) - create a a new slice using append function with length 1 & capacity of 5. y is [20].

Not true. Because there is capacity already available, it does not allocate a new array. That is the whole point of pre-allocating a slice with capacity greater than its length.

x := make([]int, 0, 5)

Yields a slice [] over an array [0, 0, 0, 0, 0].

y := append(x, 20)

Yields a slice [20] over the same array, now [20, 0, 0, 0, 0].

fmt.Println(x, y, x[:2], y[:2])

Produces four slices:

x = [] // (its length & cap haven't changed)
y = [20] // (see append explanation above)
x[:2] = [20, 0] // (first two elements of the underlying array)
y[:2] = [20, 0] // (same two elements of the same array)

This is covered in detail in the Tour of Go and in further detail on the Go blog.