Go常见的一些性能优化

2021-04-28  本文已影响0人  迪克dike

[]byte和string

转换

//强转换
func stringToBytes(s string) []byte {
   x := (*[2]uintptr)(unsafe.Pointer(&s))
   b := [3]uintptr{x[0], x[1], x[1]}
   return *(*[]byte)(unsafe.Pointer(&b))
}

func bytesToString(b []byte) string {
   return *(*string)(unsafe.Pointer(&b))
}

仅在只读场景下使用强转换

内存申请

提前预估容量

func useCap1() {
   arr := make([]int, 0, 2048)
   for i := 0; i < 2048; i++ {
      arr = append(arr, i)
   }
}

func useCap2() {
   arr := make([]int, 2048)
   for i := 0; i < 2048; i++ {
       arr[i] = i
   }
}

func noCap() {
   var arr []int
   for i := 0; i < 2048; i++ {
      arr = append(arr, i)
   }
}

Benchmark

goos: darwin
goarch: amd64
BenchmarkUseCap1-12       966577              1212 ns/op               0 B/op          0 allocs/op
BenchmarkUseCap2-12      2398420               499 ns/op               0 B/op          0 allocs/op
BenchmarkNoCap-12         192712              6016 ns/op           58616 B/op         14 allocs/op

slice扩容的主要代码,常规场景下的扩容逻辑为cap<1024时每次翻倍,cap>1024时每次增长25%,此处也可以对应上benchmark中noCap()分配在了堆上,并经过了14次扩容

newcap := old.cap
doublecap := newcap + newcap
if cap > doublecap {
   newcap = cap
} else {
   if old.len < 1024 {
      newcap = doublecap
   } else {
      // Check 0 < newcap to detect overflow
      // and prevent an infinite loop.
      for 0 < newcap && newcap < cap {
         newcap += newcap / 4
      }
      // Set newcap to the requested cap when
      // the newcap calculation overflowed.
      if newcap <= 0 {
         newcap = cap
      }
   }
}

优先在栈上分配

func BenchmarkHeap(b *testing.B) {
   m := make([]*string, 1000)
   for i := 0; i < b.N; i++ {
      for i := 0; i < 1000; i++ {
         s := "test"
         m[i] = &s
      }
   }
}

func BenchmarkStack(b *testing.B) {
   m := make([]string, 1000)
   for i := 0; i < b.N; i++ {
      for i := 0; i < 1000; i++ {
         s := "test"
         m[i] = s
      }
   }
}

Benchmark

goos: darwin
goarch: amd64
BenchmarkHeap-12           44640         23033 ns/op       16000 B/op       1000 allocs/op
BenchmarkStack-12        4650966           252 ns/op           0 B/op          0 allocs/op

Map/Slice

Map中简单结构尽量不使用指针

map[int]*int

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

func f1() {
    s := make(map[int]int, 5e7)
    for i := 0; i < 5e7; i++ {
        s[i] = i
    }
    fmt.Printf("With %T, GC took %s\n", s, gcTime())
    _ = s[0]
}

func f2() {
    s := make(map[int]*int, 5e7)
    for i := 0; i < 5e7; i++ {
        s[i] = &i
    }
    fmt.Printf("With %T, GC took %s\n", s, gcTime())
    _=s[0]
}

Output:

With map[int]int, GC took 31.956029ms
With map[int]*int, GC took 184.174966ms

不包含指针的map在gc中不需要scanObject
另外根据map的实现(关键词搜索bmap),当元素值大于128byte时,还是需要scanObject

type BigStruct struct {
   C01 int
   C02 int
   //...
   C16 int // 128byte gc scan临界点
   C17 int //136byte
 }
 
 func f3() {
   s := make(map[int]BigStruct, N)
   for i := 0; i < N; i++ {
      s[i] = BigStruct{}
   }
   fmt.Printf("With %T, GC took %s\n", s, timeGC())
   _ = s[0]
}

Output:

With map[int]main.BigStruct, GC took 1.628134832s
With map[int]main.NoBigStruct, GC took 44.708865ms

BigStruct 多了一个C17,GC时间大幅增加

对比[]*int, []int和[]BigStruct

func f4() {
    s := make([]*int, N)
    for i := 0; i < N; i++ {
        s[i] = &i
    }
    fmt.Printf("With %T, GC took %s\n", s, gcTime())
    _ = s[0]
}

func f5() {
    s := make([]int, N)
    for i := 0; i < N; i++ {
        s[i] = i
    }
    fmt.Printf("With %T, GC took %s\n", s, gcTime())
    _ = s[0]
}

func f6() {
    s := make([]BigStruct, N)
    for i := 0; i < N; i++ {
        s[i] = BigStruct{}
    }
    fmt.Printf("With %T, GC took %s\n", s, gcTime())
    _ = s[0]
}

Output:

With []*int, GC took 137.308395ms
With []int, GC took 211.862µs
With []main.BigStruct, GC took 173.504µs

slice包含指针的时候同理需要scanObject,但不包含指针时不受元素大小影响,且gc效率要比map高很多
上面的优化受到很多条条框框限制,比如map[int]string其实是包含指针的(见string定义),无法享受高效的gc,看上去不实用,但是基于此有一种应用较多的优化方式,即把大型的map结构转换为map[int]int(索引)+slice的方式,把gc压力转移到slice上(比map gc开销低),典型例子如bigcache

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