Why is there no output of the function result to the console when using Go-routines?

Question

go

As far as I understand, the main thread completes faster than the routines complete, but I don’t understand how to synchronize.

package main

import (
  "fmt"
  "net/http"
  "time"
)

func getResp()  {
  res, err := http.Get("http://ya.ru")
  if err != nil {
    fmt.Println(err)
  }
  fmt.Println(res.Status)

}

func main() {
  start := time.Now()
  for i := 0; i < 10; i++ {
    go getResp()

  }
  elapsed := time.Since(start)
  fmt.Println("Время выполнения  ", elapsed)
}

As far as I understand, the whole problem is in fmt.Println(res.Status) , I tried to remove the output from the function and call it in main via anonymous -

package main

import (
  "fmt"
  "net/http"
  "time"
)

func getResp()string  {
  res, err := http.Get("http://ya.ru")
  if err != nil {
    fmt.Println(err)
  }
  return res.Status

}

func main() {
  start := time.Now()
  for i := 0; i < 10; i++ {
    go func() {fmt.Println(getResp())}()

  }
  elapsed := time.Since(start)
  fmt.Println("Время выполнения  ", elapsed)
}

The result is the same - there is a runtime, but there are no statuses in the console. The campaign needs synchronization, but I can’t figure out how and where.

Answer 1

package main

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

var wg sync.WaitGroup

func getResp() {
  defer wg.Done()
  res, err := http.Get("http://ya.ru")
  if err != nil {
    fmt.Println(err)
  } else {
    fmt.Println(res.Status)
  }
}

func main() {
  start := time.Now()
  for i := 0; i < 10; i++ {
    wg.Add(1)
    go getResp()

  }
  wg.Wait()
  elapsed := time.Since(start)
  fmt.Println("Время выполнения  ", elapsed)
}

Answer 2

Here is an example

Answer 3

Thanks Andrey Kot , I got it more creepy

package main

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

func getResp() string {
  res, err := http.Get("http://ya.ru")
  if err != nil {
    fmt.Println(err)
    time.Sleep(1)
  }
  return res.Status
}

func main() {
  start := time.Now()
  countThread := &sync.WaitGroup{}
  for i := 0; i < 10; i++ {
    countThread.Add(1)
    go func(cT *sync.WaitGroup) {
      fmt.Println(getResp())
      cT.Done()
    }(countThread)
  }
  countThread.Wait()
  elapsed := time.Since(start)
  fmt.Println("Время выполнения  ", elapsed)
}

Answer 4

Logs? Router model? On what devices is it repeated?

Answer 5

The router's DHCP IP may be expiring.
Try to set a week, for a home network this is not critical.
Or on a Mac, set the static IP network settings

Answer 6

yes, a static IP for each MAC address in LAN, set manually (not auto) the Wi-Fi channel least clogged with neighbors, and naturally change to 8-digit new passwords for the waffle and the route's webmord itself (otherwise, maybe a bunch of neighbors got into you there, and it’s such a waffle, it gives priority to newly connected ones),
but the main thing I think is a bodyaga with a power supply for the route - do it like me, power the route through UPS (at least it’s because of the shnyaga of 220 that the route was constantly stupid. now everything is exactly already with year)