comparation about keyword "go" and without in Goroutine - go

Question

The following code logged an error:
fatal error: all goroutines are asleep - deadlock!
package main
import "fmt"
func main() {
ch := make(chan int)
ch <- 1
fmt.Println(<-ch)
}
But when I changed the code into this:
package main
import "fmt"
func assign (ch chan int) {
ch <- 1
}
func main() {
ch := make(chan int)
go assign (ch)
fmt.Println(<-ch)
}
"1" was printed out.
Then I used buffered channels:
package main
import "fmt"
func main() {
ch := make(chan int, 2)
ch <- 1
ch <- 2
fmt.Println(<-ch)
fmt.Println(<-ch)
}
"1" and "2" can also be printed out.
I'm a little confused about the situation. Thanks in advance!

Answer 1

Why the deadlock happened:
In the first code snippet you have only one main goroutine and it is blocked when you are trying to write into the channel here:
ch <- 1
Because nobody reads from the channel and the main goroutine is waiting for this to continue.
See Effective Go -> Channels
If the channel is unbuffered, the sender blocks until the receiver has received the value.
The sender is main function, the receiver is also main function.
How to avoid the deadlock:
In order to solve this, you have two options:
Option 1: make the ch channel buffered like this:
ch := make(chan int, 1) // buffer length is set to 1
From A Tour of Go
Sends to a buffered channel block only when the buffer is full.
So, you can write to the channel until the buffer is full. Then somebody has to start reading from the channel.
Option 2: write to the channel from a goroutine, like you did in the second code snippet:
func assign(ch chan int) {
ch <- 1
}
func main() {
ch := make(chan int)
go assign(ch) // does not block the main goroutine
fmt.Println(<-ch) // waiting to read from the channel
}
In this case main function will be executed until fmt.Println(<-ch) and continues as soon as it can read from the channel.

Answer 2

When you are using unbuffered channel, goroutine is blocked during write until someone does the read.
In your first snippet, there is an unbuffered channel and single goroutine (main goroutine).
So when you are trying to write:
ch <- 1
Nobody reads from the channel yet. The main goroutine is blocked and this line is never executed:
fmt.Println(<-ch)
That's why you've got the deadlock error.
In the second example, you still using unbuffered channel, which means write operation blocks the goroutine.
But by using go you are running the second goroutine.
It means even if this new goroutine will be blocked during write (in your assign function), the main goroutine will continue to work and fmt.Println(<-ch) will be executed and do the read (which in turn unblock background goroutine and assign function will finally reach the end).
To get more understanding about channels and goroutines, this snippet will give the same result (as your second snippet):
package main
import "fmt"
func print(ch chan int) {
fmt.Println(<-ch)
}
func main() {
ch := make(chan int)
go print(ch)
ch <- 1
}
When you are working with a buffered channel (third snippet), you can do N write operations without blocking goroutine (where N is the size of the buffer).
That's why in your example you did 2 writes without blocking and is able to read them later. But if your buffer is less than the count of write operations, and nobody do the read, you will fall into the same blocking issues (see the explanation of 1&2 snippets).