Go 语言用切片实现队列遇到无法理解的问题

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This topic created in 2719 days ago, the information mentioned may be changed or developed.

package main import ( "fmt" "log" "net/http" "os" "time" ) var Q []string func main() { go func() { time.Sleep(1 * time.Second) for { test() } }() Q = append(Q, "233") Q = append(Q, "233") Q = append(Q, "233") time.Sleep(3 * time.Second) Q = append(Q, "233") Q = append(Q, "233") Q = append(Q, "233") Q = append(Q, "233") Q = append(Q, "233") time.Sleep(3 * time.Second) Q = append(Q, "233") time.Sleep(3 * time.Second) Q = append(Q, "233") http.HandleFunc("/", handle) err := http.ListenAndServe(":8080", nil) if err != nil { log.Println("http port has been used.") os.Exit(-1) } } func handle(w http.ResponseWriter, r *http.Request) { defer r.Body.Close() _ = r.ParseForm() _, _ = fmt.Fprint(w, "hello") } func test() { file := queueOut() if file != "" { log.Println(file) } } func queueOut() string { res := Q[0] if len(Q) == 1 { Q[0] = "" return res } Q = Q[1:] return res }

我觉得我这样写没错啊. 但是运行的时候有几率出现

panic: runtime error: invalid memory address or nil pointer dereference [signal SIGSEGV: segmentation violation code=0x1 addr=0x0 pc=0x1058fe5] goroutine 19 [running]: fmt.(*buffer).WriteString(...) /opt/local/lib/go/src/fmt/print.go:82 fmt.(*fmt).padString(0xc000114040, 0x0, 0x3) /opt/local/lib/go/src/fmt/format.go:110 +0x9d fmt.(*fmt).fmtS(0xc000114040, 0x0, 0x3) /opt/local/lib/go/src/fmt/format.go:328 +0x61 fmt.(*pp).fmtString(0xc000114000, 0x0, 0x3, 0x76) /opt/local/lib/go/src/fmt/print.go:437 +0x122 fmt.(*pp).printArg(0xc000114000, 0x1247a20, 0xc000010050, 0x76) /opt/local/lib/go/src/fmt/print.go:671 +0x878 fmt.(*pp).doPrintln(0xc000114000, 0xc0000cffa8, 0x1, 0x1) /opt/local/lib/go/src/fmt/print.go:1146 +0xb1 fmt.Sprintln(0xc0000cffa8, 0x1, 0x1, 0x1247a20, 0x1) /opt/local/lib/go/src/fmt/print.go:271 +0x52 log.Println(0xc0000cffa8, 0x1, 0x1) /opt/local/lib/go/src/log/log.go:301 +0x3f main.test() /Users/licsber/go/src/test/main.go:52 +0xd7 main.main.func1() /Users/licsber/go/src/test/main.go:17 +0x2f created by main.main /Users/licsber/go/src/test/main.go:14 +0x39 Process finished with exit code 2

小白表示无法理解这个问题是几率出现. 每次运行不一定都有

17 replies 2018-12-07 22:24:09 +08:00

mlkr

Dec 7, 2018

要加锁吧

AlphaTr

Dec 7, 2018

slice 的线程安全问题，加锁？

Licsber

Dec 7, 2018

@mlkr
@AlphaTr 感谢我去试试

mlkr

Dec 7, 2018

package main

import (
"fmt"
"log"
"net/http"
"os"
"sync"
"time"
)

var Q []string

var lock sync.RWMutex

func main() {
go func() {
time.Sleep(1 * time.Second)
for {
test()
}
}()
Q = append(Q, "233")
Q = append(Q, "233")
Q = append(Q, "233")

time.Sleep(3 * time.Second)
Q = append(Q, "233")
Q = append(Q, "233")
Q = append(Q, "233")
Q = append(Q, "233")
Q = append(Q, "233")

time.Sleep(3 * time.Second)
Q = append(Q, "233")
time.Sleep(3 * time.Second)
Q = append(Q, "233")

http.HandleFunc("/", handle)
err := http.ListenAndServe(":8080", nil)
if err != nil {
log.Println("http port has been used.")
os.Exit(-1)
}
}
func handle(w http.ResponseWriter, r *http.Request) {
defer r.Body.Close()
_ = r.ParseForm()
_, _ = fmt.Fprint(w, "hello")

}
func test() {
file := queueOut()
if file != "" {
log.Println(file)
}

}

func queueOut() string {
// 加写锁
lock.Lock()
res := Q[0]
if len(Q) == 1 {
Q[0] = ""
return res
}
Q = Q[1:]
lock.Unlock()
return res
}

littlewing

Dec 7, 2018

有 channel 不用非要用非线程安全的 slice

另外 slice 的 append 操作是会重新分配 slice 结构的，而且你的 slice 初始 len 为 0，每次扩容时存储数据的 array 也会重新分配，问题应该就出现在这里

Licsber

Dec 7, 2018

@mlkr 好像这样子之后, 会有数据没有被加到队列里去?

Licsber
OP
Dec 7, 2018

@littlewing channel 是啥.. 刚学 go 还没看到我去看看谢谢

mritd

Dec 7, 2018 via iPhone

大哥你这个 ... channel 啊

heimeil

Dec 7, 2018

Q = Q[1:]瞬间执行多次，这个并发直接把 Q 给 cut 没了，再访问 Q 的任意下标自然就崩了。

要非得用 slice 就别直接操作 slice，另外封装两个增删方法，加上锁。

mlkr

Dec 7, 2018

@Licsber
func queueOut() string {
res := Q[0]
if len(Q) == 1 {
Q[0] = ""
return res
}
// 加写锁
lock.Lock()
Q = Q[1:]
lock.Unlock()
return res
}

AngryPanda

Dec 7, 2018 via Android

channel 没学等于没学 go

Licsber

Dec 7, 2018

@mritd 感谢成功解决问题

Licsber

Dec 7, 2018

@heimeil 对, 看了 channel 之后恍然大悟谢谢

Licsber

Dec 7, 2018

@AngryPanda 刚看到 go routine , 感谢
现在接触的 go 给我一种我一直在处理错误(err)的错觉

mritd

Dec 7, 2018

@Licsber #12 channel 是 go 的神器啊，怎么能不用呢

Licsber

Dec 7, 2018

@mlkr 感谢

Licsber

Dec 7, 2018

@mritd 刚刚看到 go 的 channel , 感谢!