#Stop runnable from evaluating expression before it's ran

1 messages · Page 1 of 1 (latest)

ebon tulip
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So I've got this player and I'm getting their location, then in the future I'm getting their new location and calculating a vector between them. Problem is, the expression that calculates the players new location is evaluated before the runnable is executed.

Entity entity = ...
Coordinates oldLocation = entity.getLocation();
delayedTasks.add(new TimedAction(2, TimeUnit.SECONDS () -> {
  Coordinates newLocation = entity.getLocation();
  Direction direction = Direction.getVector(oldLocation, newLocation);
//this is returning no direction because old location and new location are the same
}));
pine fogBOT
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<@&987246399047479336> please have a look, thanks.

pine fogBOT
#

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#

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ebon tulip
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This would be a neat Java riddle

acoustic trellis
ebon tulip
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I'm submitting the task to be delayed and I want Coordinates newLocation = entity.getLocation(); to be evaluated after that time period

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But instead it's evaluating at the time of me submitting the runnable

acoustic trellis
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Then you have a bug in the mechanism which delay the execution

ebon tulip
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No it works totally fine

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Let's say I had the following

acoustic trellis
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Then why isn't it delayed?

ebon tulip
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It is delayed

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That's not the problem

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The problem is the assignment is done when I'm creating the runnable, not when I'm executing it.

acoustic trellis
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Which assignment?

ebon tulip
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newLocation

acoustic trellis
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Then it's not delayed

ebon tulip
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It is delayed 100%

acoustic trellis
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How can you say the code in the runnable is executed when you create it, but at the same time it is delayed ?

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It's either one or the other (or both)

ebon tulip
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Okay let's say I have this

# 
int foo = 1;

int randomNumber() {
  return ...
}

Runnable runnable = () -> {
    System.out.println(foo + randomNumber());
};

runnable.run();
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random number is called on execution of the runnable

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But what about foo? does it try to access the address that foo is at when it's executed?

acoustic trellis
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foo will always be 1 in the lambda yes

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Because int is immutable

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Which isn't the case in your code

ebon tulip
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Right but is it copying the value or the reference

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because if foo was changed by some other thread, would it affect the runnable?

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Or will foo always be 1?

acoustic trellis
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It can't be changed

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It won't compile

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But if foo was a list

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And add was for example called

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Yes it would be changed

ebon tulip
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Right I understand mutability

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You're making my second guess myself because now I don't even think I know what it is I'm asking

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The delayed task processor works for everything I've thrown at it

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Minus this 1 really weird scenario

acoustic trellis
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Add a print in the lambda and outside the lambda and see if they execute at the same time

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If that's the case, then you probably have a bug

ebon tulip
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I'll try that tomorrow

acoustic trellis
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But what you said until now doesn't make sense

acoustic trellis
ebon tulip
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I'm actually gonna test it now

acoustic trellis
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Because what you are saying is that it's delayed but at the same time not delayed

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Well

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Another possible cause would be if you didn't actually edit the entity within the 2s

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You also need to check this

ebon tulip
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@acoustic trellis Nvm you're right lol

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It was running at the same time, I added a little more delay and it works just fine

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Thank you so much @acoustic trellis sorry about my nonsense but you were absolutely right