Appending one element to nil slice increases capacity by two

Question

I have a nil slice:

var s1 []int // len(s1) == 0, cap(s1) == 0 

Which I append one element to:

s2 := append(s1, 1) // len(s2) == 1, cap(s2) == 2 

Why is appending one element to a nil slice increases the capacity by 2?

Printing the slices using fmt.Printf shows the following:

[] // s1 [1] // s2 

I am also confused about why re-slicing s2[0:2] shows a zero which was not in the original slice nor appended to it:

[1,0] // s2[0:2] 

Answer 1

Go is free to provide you more capacity than you request. This improves performance by reducing the number of allocations (and possibly copying) that are required. Capacity is just the amount of space reserved before another allocation would be required.

If you append 5 elements to this slice, at least in my experiments, the capacity is 8. This shouldn't be surprising, but also shouldn't be relied on. On different platforms, or different versions of the compiler, the actual result may be different, as long as the capacity is "sufficiently large" (equal to or greater than length).

The upper index bound of a slice is defined as its capacity:

For arrays or strings, the indices are in range if 0 <= low <= high <= len(a), otherwise they are out of range. For slices, the upper index bound is the slice capacity cap(a) rather than the length. A constant index must be non-negative and representable by a value of type int; for arrays or constant strings, constant indices must also be in range. If both indices are constant, they must satisfy low <= high. If the indices are out of range at run time, a run-time panic occurs.

This is why reading past the length is not causing a panic. Even so, you shouldn't think of those zeros as part of the slice. They are indexable by the slice, but fmt.Printf(s2) would correctly not show them because they are not part of the slice. Don't subscript this way.

In general, you want to be looking at length, not capacity. Capacity mostly is readable to assist in performance optimization.