我在使用下面的代码
将byte []转换为字符串时发现了奇怪的占用堆package main
import (
"bytes"
"fmt"
"net/http"
_ "net/http/pprof"
"strings"
"time"
)
var (
c = make(chan int, 500000)
)
func main() {
go func() {
http.ListenAndServe(":8080", nil)
}()
f := func(ss []string) {
fmt.Println(ss)
time.Sleep(time.Millisecond)
<-c
}
for {
c <- 1
bs := bytes.NewBufferString("A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z").Bytes()
fmt.Println(bs) // will raise memory leak after marked as comment???
s := string(bs)
ss := strings.Split(s, ",")
go f(ss)
}
}
fmt.Println(bs)
的会逐渐耗尽记忆力。
fmt.Println(bs)
正常工作。我无法理解发生了什么?我使用version go1.9.2 darwin/amd64
答案 0 :(得分:4)
否,没有内存泄漏:
您正在使用500000
并发goroutine,您只需要限制(减少)并发goroutine的数量,例如:
c := make(chan int, runtime.NumCPU())
试试这个(看看这个编辑的结尾):
package main
import (
"bytes"
"fmt"
"runtime"
"strings"
"time"
)
func main() {
c := make(chan int, runtime.NumCPU())
for {
c <- 1
bs := bytes.NewBufferString("A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z").Bytes()
s := string(bs)
ss := strings.Split(s, ",")
go func(ss []string) {
fmt.Println(ss)
time.Sleep(time.Millisecond)
<-c
}(ss)
}
}
您的密码:
package main
import (
"bytes"
"fmt"
"net/http"
_ "net/http/pprof"
"strings"
"time"
)
var (
c = make(chan int, 500000)
)
func main() {
go func() {
http.ListenAndServe(":8080", nil)
}()
f := func(ss []string) {
fmt.Println(ss)
time.Sleep(time.Millisecond)
<-c
}
for {
c <- 1
bs := bytes.NewBufferString("A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z").Bytes()
// fmt.Println(bs) // will raise memory leak after marked as comment???
s := string(bs)
ss := strings.Split(s, ",")
go f(ss)
}
}
一段时间后达到稳定状态,甚至减少内存使用情况:
// Mem CPU time:
// 5,464,208K 0:1:20
// 5,468,208K 0:2:20
// 5,469,608K 0:3:20
// 5,469,844K 0:4:20
// 5,469,844K 0:5:20
// 5,469,848K 0:6:20
// 5,469,848K 0:7:20 fixed
// 5,469,848K 0:8:20 fixed
// 5,469,616K 0:9:20 reduced