(goroutine leaks) http.TimeoutHandler does not kill respective ServeHTTP goroutine


Timeout handler moves ServeHTTP execution on a new goroutine, but not able to kill that goroutine after the timer ends. On every request, it creates two goroutines, but ServeHTTP goroutines never kill with context.

Not able to find a way to kill goroutines.

Edit For-loop with time.Sleep function, represents huge computation which goes beyond our timer. Can replace it with any other function.

package main

import (

type api struct{}

func (a api) ServeHTTP(w http.ResponseWriter, req *http.Request) {
    // For-loop block represents huge computation and usually takes more time
    // Can replace with any code
    i := 0
    for {
        if i == 500 {
        fmt.Printf("#goroutines: %d\n", runtime.NumGoroutine())
        time.Sleep(1 * time.Second)
    _, _ = io.WriteString(w, "Hello World!")

func main() {
    var a api
    s := http.NewServeMux()
    s.Handle("/", a)
    h := http.TimeoutHandler(s, 1*time.Second, `Timeout`)

    fmt.Printf("#goroutines: %d\n", runtime.NumGoroutine())

    _ = http.ListenAndServe(":8080", h)

ServeHTTP goroutine should kill along with request context, normally which does not happen.


I found, if you do not have any way to reach to your channel then there is no way to kill or stop goroutine when it is running.

In the large computational task, you have to watch the channel on a specific interval or after specific task completion.

Answered By – Roshan Gade

Answer Checked By – Timothy Miller (GoLangFix Admin)

Leave a Reply

Your email address will not be published.