[]byte和string
转换
- 尽量避免[]byte和string的互相转换，go的string是不可变类型，标准实现中和[]byte的互转均为值拷贝
- 多数场景下都可以优先选择强转换方式进行互转
//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
- slice/map初始化尽量估计好长度，能有效减少内存分配次数，优化很明显
- 尽量规避使用append，因为需要值拷贝，且涉及到重新申请内存，可能会发生逃逸（Mac环境下测试：当append之后的slice长度大于8时会被分配到堆上）
- 如果无法预估，一般场景下可以考虑申请足够大的空间，并在场景允许的情况下优先考虑复用slice
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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