golang 并行执行 收集结果 超时控制 失败提前退出

有这么个需求
1、调用多个服务，pack1、pack2、pack3…，需要并行调用；
2、任意一个服务返回失败，提前返回失败
3、需要有超时控制不能等待太久
4、收集每个服务返回的结果，后面要做处理

怎么实现呢？

package main

import (
	"code.byted.org/gopkg/logs"
	"fmt"
	"github.com/pkg/errors"
	"time"
)

func pack1(secs time.Duration) (result string, err error) {
	fmt.Printf("waiting %v\n", secs)
	time.Sleep(secs * time.Second)
	result = fmt.Sprintf("waited for %v seconds", secs)
	//return result, errors.New("pack1 error")
	return result, nil
}

func pack2(secs time.Duration) (result string, err error) {
	fmt.Printf("waiting %v\n", secs)
	time.Sleep(secs * time.Second)
	result = fmt.Sprintf("waited for %v seconds", secs)
	return result, errors.New("pack2 error")
	//return result, nil
}

// golang parallel return functions
// 并行执行多个pack，只要有一个错误，就可以提前返回，否则需要收集多个pack结果拼接一起(reduce)
func runParallel() (string, error) {
	defer logs.Flush()
	outChan1 := make(chan *string)
	outChan2 := make(chan *string)
	go func() {
		res1, err := pack1(1)
		if err != nil {
			logs.Error("pack1 error")
			outChan1 <- nil
		} else {
			outChan1 <- &res1
		}
	}()
	go func() {
		res2, err := pack2(5)
		if err != nil {
			logs.Error("pack2 error")
			outChan2 <- nil
		} else {
			outChan2 <- &res2
		}
	}()
	var out1, out2 *string
	finishCnt := 0
	for finishCnt < 2 {
		select {
		case out1 = <-outChan1:
			if out1 == nil {
				return "", errors.New("pack1 return nil, error")
			}
			finishCnt++
		case out2 = <-outChan2:
			if out2 == nil {
				return "", errors.New("pack2 return nil, error")
			}
			finishCnt++
		case <-time.After(2 * time.Second):
			return "", errors.New("wait 2s res timeout")
		}
	}
	// 到这里就可以确保有俩结果了？
	if out1 == nil || out2 == nil {
		return "impossible", nil
	}
	return fmt.Sprintf("out1: %s, out2: %s", *out1, *out2), nil
}

func main() {
	//t0 := time.Now()
	//runNotParallel()
	t1 := time.Now()
	//fmt.Println("no parallel: ", t1.Sub(t0))
	res, err := runParallel()
	fmt.Println("res:", res, err)
	fmt.Println("parallel: ", time.Now().Sub(t1))
	time.Sleep(10 * time.Second)
}

看上去好像是ok的。测了一些case也没问题。求大神们指出是否有bug

还有一种是用syncMap实现，应该也可以

package main

import (
	"code.byted.org/gopkg/logs"
	"fmt"
	"sync"
	"time"
)

// golang parallel return functions
// 并行执行多个pack，只要有一个错误，就可以提前返回，否则需要收集多个pack结果拼接一起(reduce)
func runParallelSyncMap() (string, error) {
	defer logs.Flush()

	var (
		wgBegin   sync.WaitGroup
		synMap    sync.Map
	)

	errorChan := make(chan struct{})
	wgBegin.Add(2)
	go func() {
		defer wgBegin.Done()
		res1, err := pack1(1)
		if err != nil {
			logs.Error("pack1 error")
			errorChan <- struct{}{}
			//close(errorChan)
		} else {
			synMap.Store("pack1", res1)
		}
	}()
	go func() {
		defer wgBegin.Done()
		res2, err := pack2(2)
		if err != nil {
			logs.Error("pack2 error")
			errorChan <- struct{}{}
			//close(errorChan)// <- struct{}{}
		} else {
			synMap.Store("pack2", res2)
		}
	}()
	SyncWaitTimeout(&wgBegin, 3*time.Second, errorChan)
	// 到这里就可以确保有俩结果了？
	res := ""
	synMap.Range(func(k, v interface{}) bool {
		res += v.(string) + " +++  "
		return true
	})

	return fmt.Sprintf("out: %s", res), nil
}

func main() {
	//t0 := time.Now()
	//runNotParallel()
	t1 := time.Now()
	//fmt.Println("no parallel: ", t1.Sub(t0))
	res, err := runParallelSyncMap()
	fmt.Println("res:", res, err)
	fmt.Println("parallel: ", time.Now().Sub(t1))
	time.Sleep(10 * time.Second)
}

func SyncWaitTimeout(wg *sync.WaitGroup, timeout time.Duration, errorChan chan struct{}) bool {
	defer logs.Flush()
	finish := make(chan struct{})
	go func() {
		defer close(finish)
		wg.Wait()
	}()
	select {
	case <-finish:
		return true
	case <-errorChan:
		logs.Error("errorChan")
		return false
	case <-time.After(timeout):
		logs.Error("timeout")
		return false
	}
}