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Golang的GC性能优化技巧

slice 预先分配内存

slice的容量小于1024时,容量是按照2倍大小增长的。当容量大于1024,增长的容量是原来的1.25倍。看下面的离例子:

func appendOne(num int) []int {
	var res []int
	for i := 0; i < num; i++ {
		res = append(res, i)
	}
	return res
}

func appendMany(num int) []int {
	res := make([]int, 0, num)
	for i := 0; i < num; i++ {
		res = append(res, i)
	}
	return res
}
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函数appendOne没有指定初始容量大小,appendMany指定了初始的容量大小。进行一下benchmark测试:

func BenchmarkAppendOne(b *testing.B) {
	num := 10000
	for i := 0; i < b.N; i++ {
		_ = appendOne(num)
	}
}

func BenchmarkAppendMany(b *testing.B) {
	num := 10000
	for i := 0; i < b.N; i++ {
		_ = appendMany(num)
	}
}
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运行测试

$ go test -bench=. -benchmem                                                      
goos: darwin
goarch: amd64
pkg: com.learn/gormLearn/gc_gc
BenchmarkAppendOne-4               23163             50675 ns/op          386296 B/op         20 allocs/op
BenchmarkAppendMany-4              96781             12241 ns/op           81920 B/op          1 allocs/op
PASS
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可以看到,AppendMany每次进行操作进行1次的内存分配,每次内存分配,分配了81920 B, 每次操作耗时12241 ns,这些指标都好于AppendOne。 一次性分配需要的内存大小,slice不需要在扩大底层数组时进行内存分配,旧的底层数据依然能够复用,这显然减少了GC的压力。

同样地,新建map也可以指定大小。

func makeMap(num int){
    m := make(map[int]int,num)
    for i:=0;i<len(num);i++{
        m[i]=i
    }
}
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这可以减少内存拷贝的开销,也可以减少rehash开销。

map中保存值,而不是指针,使用分段map

看下面的列子,map分别保存指针,值。

func timeGC() time.Duration {
	start := time.Now()
	runtime.GC()
	return time.Since(start)
}

func mapPointer(num int) {
	m := make(map[int]*int, num)
	for i := 0; i < num; i++ {
		m[i] = &i
	}
	runtime.GC()
	fmt.Printf("With %T, GC took %s\n", m, timeGC())
	_ = m[0]
}

func mapValue(num int) {
	m := make(map[int]int, num)
	for i := 0; i < num; i++ {
		m[i] = i
	}
	runtime.GC()
	fmt.Printf("With %T, GC took %s\n", m, timeGC())
	_ = m[0]
}

func mapPointerShard(num int) {
	shards := make([]map[int]*int, 100)
	for i := range shards {
		shards[i] = make(map[int]*int)
	}
	for i := 0; i < num; i++ {
		shards[i%100][i] = &i
	}
	runtime.GC()
	fmt.Printf("With map shards (%T), GC took %s\n", shards, timeGC())
	_ = shards[0][0]
}

func mapValueShard(num int) {
	shards := make([]map[int]int, 100)
	for i := range shards {
		shards[i] = make(map[int]int)
	}
	for i := 0; i < num; i++ {
		shards[i%100][i] = i
	}
	runtime.GC()
	fmt.Printf("With map shards (%T), GC took %s\n", shards, timeGC())
	_ = shards[0][0]
}

const N = 5e7 // 5000w

func BenchmarkMapPointer(b *testing.B) {
	mapPointer(N)
}

func BenchmarkMapValue(b *testing.B) {
	mapValue(N)
}

func BenchmarkMapPointerShard(b *testing.B) {
	mapPointerShard(N)
}

func BenchmarkMapValueShard(b *testing.B) {
	mapValueShard(N)
}
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运行

$ go test -bench=^BenchmarkMapPointer$ -benchmem
With map[int]*int, GC took 545.139836ms
goos: darwin
goarch: amd64
pkg: com.learn/gormLearn/gc_gc
BenchmarkMapPointer-4                  1        9532798100 ns/op        1387850488 B/op   724960 allocs/op

$ go test -bench=^BenchmarkMapPointerShard$ -benchmem
With map shards ([]map[int]*int), GC took 688.39764ms
goos: darwin
goarch: amd64
pkg: com.learn/gormLearn/gc_gc
BenchmarkMapPointerShard-4             1        20670458639 ns/op       4286763416 B/op  1901279 allocs/op

$ go test -bench=^BenchmarkMapValueShard$ -benchmem
With map shards ([]map[int]int), GC took 1.965519ms
goos: darwin
goarch: amd64
pkg: com.learn/gormLearn/gc_gc
BenchmarkMapValueShard-4               1        16190847776 ns/op       4385268936 B/op  1918445 allocs/op

$ go test -bench=^BenchmarkMapValue$ -benchmem 
With map[int]int, GC took 22.993926ms
goos: darwin
goarch: amd64
pkg: com.learn/gormLearn/gc_gc
BenchmarkMapValue-4            1        8253025035 ns/op        1444338752 B/op   724512 allocs/op
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可以看到,使用分段的,保存值的map的GC耗时最小。 加上GODEBUG=gctrace=1分析GC轨迹:

$ GODEBUG=gctrace=1 go test -bench=^BenchmarkMapPointer$ -benchmem
...
gc 3 @0.130s 19%: 0.006+424+0.013 ms clock, 0.027+0.18/424/848+0.055 ms cpu, 1224->1224->1224 MB, 1225 MB goal, 4 P
gc 4 @9.410s 2%: 0.005+543+0.002 ms clock, 0.022+0/543/1628+0.011 ms cpu, 1325->1325->1323 MB, 2448 MB goal, 4 P (forced)
gc 5 @9.957s 3%: 0.003+547+0.003 ms clock, 0.013+0/547/1631+0.013 ms cpu, 1323->1323->1323 MB, 2647 MB goal, 4 P (forced)
With map[int]*int, GC took 550.40821ms

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为了理解打印的日志,我们要理解gctrace, 以0.013+0/547/1631+0.013 ms cpu为例子,GC分为三个阶段。

  • Mark Prepare (STW)0.013表示标记阶段的全局暂停(stop the wrold)时间。
  • Marking0/547/1631, 0表示mutator assist的耗时,5471631都是标记GC的耗时。
  • Mark Termination (STW)0.013表示标记结束阶段的全局暂停(stop the wrold)时间。
$ GODEBUG=gctrace=1 go test -bench=^BenchmarkMapValue$ -benchmem
...
gc 3 @0.018s 0%: 0.005+0.14+0.015 ms clock, 0.021+0.054/0.020/0.19+0.060 ms cpu, 1224->1224->1224 MB, 1225 MB goal, 4 P
gc 4 @8.334s 0%: 0.006+21+0.003 ms clock, 0.027+0/6.4/21+0.013 ms cpu, 1379->1379->1334 MB, 2448 MB goal, 4 P (forced)
gc 5 @8.358s 0%: 0.003+19+0.003 ms clock, 0.014+0/5.0/20+0.015 ms cpu, 1334->1334->1334 MB, 2668 MB goal, 4 P (forced)
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可以看到,map保存值比保存指针的耗时少,主要是在GC的标记阶段耗时更少。

string与[]byte的转换

Golang中,string从设计上是不可变(immutable)的。因此,string[]byte的类型转换,都是产生一份新的副本。

func Example() {
	s := "Hello,world"
	b := []byte(s)
}
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如果确定转换的string/[]byte不会被修改,可以进行直接的转换,这样不会生成原有变量的副本。新的变量共享底层的数据指针。

func String2Bytes(s string) []byte {
	stringHeader := (*reflect.StringHeader)(unsafe.Pointer(&s))
	bh := reflect.SliceHeader{
		Data: stringHeader.Data,
		Len:  stringHeader.Len,
		Cap:  stringHeader.Len,
	}
	return *(*[]byte)(unsafe.Pointer(&bh))
}

func Bytes2String(b []byte) string {
	sliceHeader
	sh := reflect.StringHeader{
		Data: sliceHeader.Data,
		Len:  sliceHeader.Len,
	}
	return *(*string)(unsafe.Pointer(&sh))
}
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函数返回值使用值,不使用指针

对占用空间少,频繁分配的函数,如果函数返回指针,会带来内存逃逸,使得原来可以分配在栈(stack)上的内存,需要分配在堆(heap)上。在栈上进行小对象拷贝的性能很好,比分配对象在堆上要好得多。 看下面的例子,2个函数分别返回值,和指针。

type S struct {
	a, b, c int64
	d, e, f string
	g, h, i float64
}

func byCopy() S {
	return S{
		a: 1, b: 1, c: 1,
		e: "lyp", f: "lyp",
		g: 1.0, h: 1.0, i: 1.0,
	}
}

func byPointer() *S {
	return &S{
		a: 1, b: 1, c: 1,
		e: "lyp", f: "lyp",
		g: 1.0, h: 1.0, i: 1.0,
	}
}
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benchmark函数

func BenchmarkMemoryStack(b *testing.B) {
	var s S

	f, err := os.Create("stack.out")
	if err != nil {
		panic(err)
	}
	defer f.Close()

	err = trace.Start(f)
	if err != nil {
		panic(err)
	}

	for i := 0; i < b.N; i++ {
		s = byCopy()
	}

	trace.Stop()

	b.StopTimer()
	_ = fmt.Sprintf("%v", s.a)
}

func BenchmarkMemoryHeap(b *testing.B) {
	var s *S

	f, err := os.Create("heap.out")
	if err != nil {
		panic(err)
	}
	defer f.Close()

	err = trace.Start(f)
	if err != nil {
		panic(err)
	}

	for i := 0; i < b.N; i++ {
		s = byPointer()
	}

	trace.Stop()

	b.StopTimer()
	_ = fmt.Sprintf("%v", s.a)
}
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运行

 go test ./... -bench=BenchmarkMemoryHeap -benchmem -run=^$ -count=10           
goos: darwin
goarch: amd64
pkg: com.learn/gormLearn/gc_gc
BenchmarkMemoryHeap-4           19625536                53.0 ns/op            96 B/op          1 allocs/op

go test ./... -bench=BenchmarkMemoryStack -benchmem -run=^$ -count=10                                   
goos: darwin
goarch: amd64
pkg: com.learn/gormLearn/gc_gc
BenchmarkMemoryStack-4          163253341                7.22 ns/op            0 B/op          0 allocs/op
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可以看到,栈分配(返回值)的耗时是7.22 ns/op,而堆分配(返回指针)的耗时是53.0 ns/op

使用struct{}优化

Golang中,没有集合(set)。如果要实现一个集合,可以使用struct{}作为值。

func assign(num int) {
	m := make(map[int]bool, num)
	for i := 0; i < num; i++ {
		m[i] = true
	}
}

func assignStruct(num int) {
	m := make(map[int]struct{}, num)
	for i := 0; i < num; i++ {
		m[i] = struct{}{}
	}
}
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struct{}经过编译器特殊优化,指向同一个内存地址(runtime.zerobase),不占用空间。

GC分析的工具

  • go tool pprof
  • go tool trace
  • go build -gcflags=”-m”
  • GODEBUG=”gctrace=1”

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