Synchronous method call

loma

I am a GDScript beginner. Here's my scenario:

func m1():
	if(test):
		yield(animate(), 'completed')

func m2():
	m1()
	print('done')

I want print to be called after m1 is finished. But if I put m1 call in yield, I get error if condition test is not fulfilled.

What's the best approach for such situations.

fire7side

Can't you do something like this:

var test = true

func _ready():
	if(test):
		try_timer()

func try_timer():
	$Timer.start()
	yield($Timer,"timeout")
	print("finished")

loma

@fire7side Thank you for your response. I don't think this your approach would address my problem. Simply put the problem is: if I want to call a method that may or may not do asynchronous tasks like tween.start(), how do I wait for it's completion. I am assuming, you wait by making a call to yield(fn, 'completed').

fire7side

I'm pretty sure tween gives a completed signal, and that's how you wait for it, but when you have a branch like that with a yield that may or may not be called so the function might wait or not, I'm not sure the compiler will like that. I haven't really done it though. I always call a function that has the yield in it that does the waiting in the branch instead. That's what the example was for. It may work though, I just don't do it that way.