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Golang
Why does passing a WaitGroup by value in Go lead to a deadlock, and how can it be resolved?
Why does passing a WaitGroup by value in Go lead to a deadlock, and how can it be resolved?
Deadlock with Go Channels: An Issue of Variable Scope
In a Golang program, channels facilitate communication between goroutines. However, misusing channels can lead to deadlocks, as demonstrated in the code below:
<br>package main</p> <p>import (</p> <div class="code" style="position:relative; padding:0px; margin:0px;"><div class="code" style="position:relative; padding:0px; margin:0px;"><pre class="brush:php;toolbar:false">"fmt" "sync"
)
func push(c chan int, wg sync.WaitGroup) {
for i := 0; i < 5; i++ {
c <- i
}
wg.Done()}
func pull(c chan int, wg sync.WaitGroup) {
for i := 0; i < 5; i++ {
result, ok := <-c
fmt.Println(result, ok)
}
wg.Done()}
func main() {
var wg sync.WaitGroup wg.Add(2) c := make(chan int) go push(c, wg) go pull(c, wg) wg.Wait() // Block the main thread until goroutines complete
}
When running this program, you might encounter the following error:
fatal error: all goroutines are asleep - deadlock!
To understand why this deadlock occurs, let's delve into the code:
- main creates a WaitGroup, a channel c, and goroutines for push and pull operations.
- The push and pull functions use the WaitGroup to synchronize their execution.
- The push function sends values to c in a loop and signals its completion by invoking wg.Done().
- The pull function receives values from c and prints them. It also signals completion with wg.Done().
The problem lies in how the WaitGroup is passed to the goroutines. When a value is passed without an ampersand (&), it is passed by value and not by reference. In this case, a copy of the WaitGroup is created for each goroutine.
As a result, when each goroutine calls wg.Done(), it modifies its local copy of the WaitGroup. Since the main thread waits until wg indicates that all goroutines are finished, it waits indefinitely because neither goroutine updates the original WaitGroup. This leads to a deadlock.
To resolve this issue, we need to pass the WaitGroup by reference. This ensures that both goroutines modify the same instance of the WaitGroup and correctly signal their completion to the main thread.
Here's a revised version of the code with the correction:
<br>package main</p> <p>import (</p> <div class="code" style="position:relative; padding:0px; margin:0px;"><div class="code" style="position:relative; padding:0px; margin:0px;"><pre class="brush:php;toolbar:false">"fmt" "sync"
)
func push(c chan int, wg *sync.WaitGroup) {
for i := 0; i < 5; i++ {
c <- i
}
wg.Done()}
func pull(c chan int, wg *sync.WaitGroup) {
for i := 0; i < 5; i++ {
result, ok := <-c
fmt.Println(result, ok)
}
wg.Done()}
func main() {
var wg sync.WaitGroup wg.Add(2) c := make(chan int) go push(c, &wg) // Pass the WaitGroup by reference using the ampersand go pull(c, &wg) // Pass the WaitGroup by reference using the ampersand wg.Wait()
}
By passing the WaitGroup by reference, we ensure that the main thread can correctly determine when both goroutines have completed their tasks, thus avoiding the deadlock.
The above is the detailed content of Why does passing a WaitGroup by value in Go lead to a deadlock, and how can it be resolved?. For more information, please follow other related articles on the PHP Chinese website!
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