At first glance, this code looks ok to me. Here's the code:
func main() {
mychan := make(chan string)
text := "My Text"
mychan <- text // DON'T DO THIS!!! Passing chan values should be within routines
fmt.Println(<-mychan)
}
But it will return deadlock errors. Something like this:
fatal error: all goroutines are asleep - deadlock!
Then, I do something like this to overcome this error:
func main() {
mychan := make(chan string)
go func() {
text := "My Text"
mychan <- text
}()
fmt.Println(<-mychan)
}
But when I try to do multiple routines:
func main() {
mychan := make(chan string)
go func() {
text := "My Text"
mychan <- text
}()
go func() {
text := "My Text 2"
mychan <- text
}()
fmt.Println(<-mychan)
fmt.Println(<-mychan)
}
It will output the value of My Text and My Text 2 without following any order. The reason behind this is that go-routine executes functions concurrently. We really don't know which comes first and which comes last.
To solve this either we need to declare another variable for chan string or use a WaitGroup.
Let's see what it looks like when we introduce a new variable.
func main() {
mychan := make(chan string)
mychan2 := make(chan string) // INTRODUCE NEW VARIABLE
go func() {
text := "My Text"
mychan <- text
}()
go func() {
text := "My Text 2"
mychan2 <- text
}()
fmt.Println(<-mychan)
fmt.Println(<-mychan2)
}
And there's another way we can solve this by using WaitGroup. Here's an example:
func main() {
mychan := make(chan string)
var wg sync.WaitGroup
wg.Add(1)
go func() {
text := "My Text"
wg.Done() // IMPORTANT: This must write before sending value to the channel
mychan <- text
}()
wg.Wait() // We are going to wait until the first routine function to finish
go func() {
text := "My Text 2"
mychan <- text
}()
fmt.Println(<-mychan)
fmt.Println(<-mychan)
}