golang range for循环中如何正确的给goroutine传参

Posted 2021-02-12 waken-captain

1.code example

 公共方法

func testDomain(ii string) {
	time.Sleep(time.Second * 4)
	fmt.Printf("pid: %d___point addr: %d___%s 
", GoID(), &ii, ii)
}

func GoID() int {
	var buf [64]byte
	n := runtime.Stack(buf[:], false)
	idField := strings.Fields(strings.TrimPrefix(string(buf[:n]), "goroutine "))[0]
	id, err := strconv.Atoi(idField)
	if err != nil {
		panic(fmt.Sprintf("cannot get goroutine id: %v", err))
	}
	return id
}

　　

2. 错误示范

var a []string
for i := 1; i < 5; i++ {
   a = append(a, fmt.Sprintf("%d", i))
}

	for _, i := range a {
		fmt.Printf("-----%s---
", i)

		go func() {
			time.Sleep(time.Second * 4)
			testDomain(i)
		}()
	
	}

　打印发现i每次地址都是同一个

    协助每次先阻塞4秒

　 4秒后 i的值是4， 这是协程中的方法testDomain开始工作，将i的值传给自己的形参

3. 正确示范

	for _, i := range a {
		fmt.Printf("-----%s---
", i)
		go func(a string) {
			//time.Sleep(time.Second * 4)
			testDomain(a)
		}(i)
	}

　这种操作会先将i的值传递给形参a，i的变化不会对testDomain方法的执行产生影响

 

4. 完整代码

package main

import (
	"fmt"
	"runtime"
	"strconv"
	"strings"
	"time"
)

func main() {
	var a []string
	for i := 1; i < 5; i++ {
		a = append(a, fmt.Sprintf("%d", i))
	}
	for _, i := range a {
		fmt.Printf("-----%s---
", i)
		go func(a string) {
			//time.Sleep(time.Second * 4)
			testDomain(a)
		}(i)

		go func() {
			time.Sleep(time.Second * 4)
			testDomain(i)
		}()
		fmt.Println(&i)
		time.Sleep(time.Second * 1)
	}

	time.Sleep(100 * time.Second)
}

func testDomain(ii string) {
	time.Sleep(time.Second * 4)
	fmt.Printf("pid: %d___point addr: %d___%s 
", GoID(), &ii, ii)
}

func GoID() int {
	var buf [64]byte
	n := runtime.Stack(buf[:], false)
	idField := strings.Fields(strings.TrimPrefix(string(buf[:n]), "goroutine "))[0]
	id, err := strconv.Atoi(idField)
	if err != nil {
		panic(fmt.Sprintf("cannot get goroutine id: %v", err))
	}
	return id
}