testing - 去基准和 gc : B/op alloc/op

Question

基准代码:

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

结果:

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

问题:

• 40 B/op 从何而来？ (非常感谢任何追踪方式+说明)
• 怎么可能有 40 个 B/op 而有 0 个分配？
• 哪个会影响 GC，如何影响？ (B/op 或 allocs/op)
• 真的有可能使用 append 得到 0 B/op 吗？

Answer 1

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.

对于您的示例，平均而言，每个操作分配 [40, 41) 字节以在必要时增加 slice 的容量。使用摊销常数时间算法增加容量:直到 len 1024 增加到上限的 2 倍，然后增加到上限的 1.25 倍。平均而言，每个操作有 [0, 1) 个分配。

例如，

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))
}

输出:

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

对于 op = 50000000，

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

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

阅读:

Go Slices: usage and internals

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

'append' complexity

要使追加的 B/op(和 allocs/op)为零，请在追加之前分配一个具有足够容量的 slice 。

例如，使用 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))
}

输出:

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

请注意基准 CPU 时间从大约 26.6 ns/op 减少到大约 11.7 ns/op。