Because you have an unbuffered channel, and you can only send a value on an unbuffered channel without blocking if there is another goroutine which is ready to receive from it.
Since you only have 1 goroutine, it gets blocked. Solution is simple: launch your Quack.quack()
method in a new goroutine:
go self.quack(ch)
Then the output (try it on the Go Playground):
mockQuack start
mockQuack done
true
Another option is to not launch a new goroutine but make a buffered channel, so it can hold some values without any receivers ready to receive from it:
ch := make(chan bool, 1) // buffered channel, buffer for 1 value
This creates a channel which is capable of "storing" one value without any receivers ready to receive it. A second send on the channel would also block, unless the value is received from it first (or a receiver ready to receive a value from it).
Try this buffered channel version on the Go Playground.
Relevant section from the spec: Send statements:
Both the channel and the value expression are evaluated before communication begins. Communication blocks until the send can proceed. A send on an unbuffered channel can proceed if a receiver is ready. A send on a buffered channel can proceed if there is room in the buffer. A send on a closed channel proceeds by causing a run-time panic. A send on a nil channel blocks forever.
Notes:
Based on the received value you print true
or false
. This can be done with a single line, without the if
statement:
fmt.Println(b)
You can even get rid of the b
local variable, and print the received value right away:
fmt.Println(<-ch)
Also I assume you used channels because you wanted to play with them, but in your case mockQuack()
could simply return the bool
value, without the use of channels.