I have a subroutine (that I call with a link call node) which doesn't have a link out node (i.e. "SendToTelegram"). How can I avoid the timeout of the link call node?
if you're not expecting a return for your link call just use a link out and in only without a link-call. A link call expects a return
No Return Expected
Return Expected
Ad for those times when a naughty node swallows the reply (resulting in a timeout) you can catch the timeout error and act accordingly.
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@Steve-Mcl I noticed that a subroutine cannot be part of a subflow. I am getting the error: Missing return node information
A link call can call out of a subflow to an external subroutine but it makes no sense to call into a subroutine.
Think for a second what a subflow does: it generates n copies of its self - so which link-in would a link-call call to?
Additionally, link-call does work internally in a subflow when calling a link-in that exists in the same subflow.
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Got it. Thanks
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