go benchmark and gc: B/op alloc/op

benchmark code:

func BenchmarkSth(b *testing.B) {
	var x []int
	b.ResetTimer()
	for i := 0; i < b.N; i++ {
		x = append(x, i)
	}
}

result:

BenchmarkSth-4    50000000    20.7 ns/op    40 B/op    0 allocs/op 

question/s:

• where did 40 B/op come from? (any way of tracing + instructions is greatly appreciated)
• how is it possible to have 40 B/op while having 0 allocs?
• which one affects GC and how? (B/op or allocs/op)
• is it really possible to have 0 B/op using append?

> The Go Programming Language Specification
>
> Appending to and copying slices
>
> The variadic function append appends zero or more values x to s of
> type S, which must be a slice type, and returns the resulting slice,
> also of type S.
>
> append(s S, x ...T) S // T is the element type of S
>
> If the capacity of s is not large enough to fit the additional values,
> append allocates a new, sufficiently large underlying array that fits
> both the existing slice elements and the additional values. Otherwise,
> append re-uses the underlying array.

For your example, on average, [40, 41) bytes per operation are allocated to increase the capacity of the slice when necessary. The capacity is increased using an amortized constant time algorithm: up to len 1024 increase to 2 times cap then increase to 1.25 times cap. On average, there are [0, 1) allocations per operation.

For example,

func BenchmarkMem(b *testing.B) {
	b.ReportAllocs()
	var x []int64
	var a, ac int64
	b.ResetTimer()
	for i := 0; i < b.N; i++ {
		c := cap(x)
		x = append(x, int64(i))
		if cap(x) != c {
			a++
			ac += int64(cap(x))
		}
	}
	b.StopTimer()
	sizeInt64 := int64(8)
	B := ac * sizeInt64 // bytes
	b.Log("op", b.N, "B", B, "alloc", a, "lx", len(x), "cx", cap(x))
}

Output:

BenchmarkMem-4   	50000000	        26.6 ns/op	      40 B/op	       0 allocs/op
--- BENCH: BenchmarkMem-4
	bench_test.go:32: op 1 B 8 alloc 1 lx 1 cx 1
	bench_test.go:32: op 100 B 2040 alloc 8 lx 100 cx 128
	bench_test.go:32: op 10000 B 386296 alloc 20 lx 10000 cx 12288
	bench_test.go:32: op 1000000 B 45188344 alloc 40 lx 1000000 cx 1136640
	bench_test.go:32: op 50000000 B 2021098744 alloc 57 lx 50000000 cx 50539520

For op = 50000000,

B/op = floor(2021098744 / 50000000) = floor(40.421974888) = 40

allocs/op = floor(57 / 50000000) = floor(0.00000114) = 0

Read:

Go Slices: usage and internals

Arrays, slices (and strings): The mechanics of 'append'

'append' complexity

To have zero B/op (and zero allocs/op) for append, allocate a slice with sufficient capacity before appending.

For example, with var x = make([]int64, 0, b.N),

func BenchmarkZero(b *testing.B) {
	b.ReportAllocs()
	var x = make([]int64, 0, b.N)
	var a, ac int64
	b.ResetTimer()
	for i := 0; i < b.N; i++ {
		c := cap(x)
		x = append(x, int64(i))
		if cap(x) != c {
			a++
			ac += int64(cap(x))
		}
	}
	b.StopTimer()
	sizeInt64 := int64(8)
	B := ac * sizeInt64 // bytes
	b.Log("op", b.N, "B", B, "alloc", a, "lx", len(x), "cx", cap(x))
}

Output:

BenchmarkZero-4   	100000000	        11.7 ns/op	       0 B/op	       0 allocs/op
--- BENCH: BenchmarkZero-4
	bench_test.go:51: op 1 B 0 alloc 0 lx 1 cx 1
	bench_test.go:51: op 100 B 0 alloc 0 lx 100 cx 100
	bench_test.go:51: op 10000 B 0 alloc 0 lx 10000 cx 10000
	bench_test.go:51: op 1000000 B 0 alloc 0 lx 1000000 cx 1000000
	bench_test.go:51: op 100000000 B 0 alloc 0 lx 100000000 cx 100000000

Note the reduction in benchmark CPU time from around 26.6 ns/op to around 11.7 ns/op.