JuniorTheoryCommon

What does the program print?

package main

import "fmt"
import "sync"

func main() {
    var wg, mu, n = sync.WaitGroup{}, sync.Mutex{}, 0
    wg.Go(func() { mu.Lock(); n++; mu.Unlock() })
    wg.Go(func() { mu.Lock(); n++; mu.Unlock() })
    wg.Wait()
    fmt.Println(n)
}

Expecting 1: mu.Lock lets only one goroutine run n++ at a time, so the second one sees 1 and no increment is lost.
Expecting a deadlock: the second Lock only waits until the first goroutine calls Unlock, then it goes on.

What could the program print without mu.Lock and mu.Unlock?
Why must wg.Wait() come before fmt.Println(n)?

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JuniorTheoryCommon

What does the program print?

package main

import "fmt"
import "sync"

func main() {
    var wg, mu, n = sync.WaitGroup{}, sync.Mutex{}, 0
    wg.Go(func() { mu.Lock(); n++; mu.Unlock() })
    wg.Go(func() { mu.Lock(); n++; mu.Unlock() })
    wg.Wait()
    fmt.Println(n)
}

Expecting 1: mu.Lock lets only one goroutine run n++ at a time, so the second one sees 1 and no increment is lost.
Expecting a deadlock: the second Lock only waits until the first goroutine calls Unlock, then it goes on.

What could the program print without mu.Lock and mu.Unlock?
Why must wg.Wait() come before fmt.Println(n)?

Back to questions Open topic in roadmap →Answer it yourself — Quiz