Code doesn't work when put in a function

kstcold

func _ready():
	var time_then = OS.get_unix_time()
	$Timer.start(2)
	yield($Timer, "timeout")
	print(OS.get_unix_time() - time_then)

This prints 2 and there will be a delay

However, there is no delay here:

func _ready():
	var time_then = OS.get_unix_time()
	add_delay(2)
	print(OS.get_unix_time() - time_then)

func add_delay(duration):
	$Timer.start(duration)
	yield($Timer, "timeout")

This prints 0 and there is no delay

cybereality

Just never use yield. Use Timers.

fire7side

Yeah, once you break it up like that, you want a function that is connected and waits for the timeout signal. The compiler treats that type of function with a yield in it differently so you have to include the code that comes after like you did in the first function or you could call another function afterwards.

xyz

@kstcold said: This prints 0 and there is no delay

Of course because when you call yield() execution immediately returns from the current function back to the caller, which is in this case _ready(). If you put time measurement inside add_delay() it will measure/print it correctly.

If you don't fully understand why is this so, better stay away from yields, as already said. There are simpler alternatives.