How to Block Forever in Go

Jun 5, 2016 · 3 minute read · Comments
Go

Let me count the ways

There seems to be quite a few ways to block forever in Go. This post is part code golf and part practical advice. All of the code below is simply to find a way to block the current goroutine from progressing while still allowing others to continue. The versions that block forever will actually cause a panic saying that all goroutines are asleep, but the ones that “busy block” will just time out on the playground.

Why Not an Infinite Loop?

An infinite loop will simply use 100% of a CPU an not allow the Go runtime to schedule any other work on that core. All of the solutions below will either block forever or very quickly yield the processor for other work.

(thanks @bradleyfalzon)

Empty Select

This function tries to select the first case that returns, so it will happily wait forever for nothing to return.

func blockForever() {
    select{ }
}

Waiting on Itself

A sync.WaitGroup is a way to coordinate multiple goroutines by reporting completion. Since there’s nobody to report completion, it never returns.

import "sync"

func blockForever() {
    wg := sync.WaitGroup{}
    wg.Add(1)
    wg.Wait()
}

Double Locking

A sync.Mutex provides exclusive use of another associated resource. What if it’s requested twice? Oops, blocked.

import "sync"

func blockForever() {
    m := sync.Mutex{}
    m.Lock()
    m.Lock()
}

The same trick works with a sync.RWMutex that has been RLock()’d.

Reading from an Empty Channel

Empty unbuffered channels will block until there’s something to receive.

func blockForever() {
    c := make(chan struct{})
    <-c
}

This also works with a nil channel not made with make() (H/T: @cgilmour on the Gophers Slack).

“Busy Blocking”

This method technically doesn’t block, it just constantly defers to other work if it’s available.

import "runtime"

func blockForever() {
    for {
        runtime.Gosched()
    }
}

The Decomposed Loop

Same as before, not technically blocking but looping and deferring to other work.

import "runtime"

func blockForever() {
    foo: runtime.Gosched()
    goto foo
}

Shaking my Own Hand

This is a bit like shaking your own hand. This function will continually send a useless message to itself until the end of time. The channel send operations are opportunities for the runtime to schedule other goroutines, so this method would not monopolize a single processor.

func blockForever() {
    c := make(chan struct{}, 1)
    for {
        select {
        case <-c:
        case c <- struct{}{}:
        }
    }
}

Sleeping for a Looooong Time

Sleeping to the maximum time allowed takes quite a long time, roughly 292.4 years. Likely, your code won’t run that long, so I’d consider this equivalent to select{}.

import (
    "math"
    "time"
)

func blockForever() {
    <-time.After(time.Duration(math.MaxInt64))
}

This comes from @hydroflame (on the Gophers Slack) of Lux fame.

Are These Useful?

The first one is likely the only one you’ll ever need. For example, if you have a single server with multiple http listeners on different ports, you can run them all as separate goroutines in your main() and just block forever with a select{}. It’s likely that any failure of a server ought to end your program, so having errors panic() will end the process.

If for some reason you don’t want your program to end on a single server failing (or stopping), you can create a *sync.WaitGroup to pass in to each wrapper goroutine and wait on it before ending the main() function.

import (
    "log"
    "net/http"
    "sync"
)

func main() {
    wg := &sync.WaitGroup{}
    wg.Add(4)
    
    for i := 1111; i < 1115; i++ {
        go func(port int, wg *sync.WaitGroup) {
            err := http.ListenAndServe(fmt.Sprintf(":%d", port), someHandlerFunction)
            log.Print(err)
            wg.Done()
        }(i, wg)
    }
    
    wg.Wait()
}

The rest of the methods are just for fun. if you have any other methods that you know of that will not monopolize a CPU, leave a comment and let me know. Enjoy :)

Scott Mansfield