How to do IF (a or b or c) == true?

Question

Doing FSM and sometimes you want the same logic to be applied for multiple states, for instance, aiming a weapon should be allowed when the player is idle, walking or running.

For such common situation normally I would do, as you guessed:

#AIM WEAPON - logic
func Something_Foobar() -> void:
  if STATE == "IDLE" or STATE == "WALK" or STATE == "RUN": pass
    # Allow aim weapon code check.
      # Allow firing weapon.

How I would go about on doing them like this: if STATE == ("IDLE" or "WALK" or "RUN"): or better even if STATE == ("IDLE","WALK","RUN"): or the inverse if ("IDLE" or "WALK" or "RUN") in STATE:

I would avoid using (arrays or dictionaries or for) loops just for this.

Thanks for reading,

Answer 1

Arrays seem to be the correct answer here

if STATE in ["IDLE","WALK","RUN"]:

Comments

Oh~ just remembered. Since you're doing FSM you can also use match

match STATE:
    "IDLE", "WALK", "RUN":
        # Allow aim weapon code check.
        continue
    "FIRE":
        #allow reload check
        break
    "WALK":
        #do only walk related stuff
         break

Pay attention to the "WALK" STATE being used multiple times
And this can also be all placed in one line if you're one of those fanatics

Answer 2

In addition to the array suggested by Wakatta, a second alternative would be using match, especially if you've got more than one group with specific code:

match STATE:
  "IDLE", "WALK", "RUN":
    # Code when standing on the ground
  "JUMP", "FALL":
    # Code when in the air
  "SWIM", "FLOAT":
    # Code when in a liquid

Answer 3

The previous answers are great but I've got two more approaches:

1) OOP with inheritance:

This can handle transitioning states and general behaviour without branching at all.

Tutorials here: https://www.gdquest.com/tutorial/godot/design-patterns/finite-state-machine/ and here https://docs.godotengine.org/en/stable/tutorials/misc/state_design_pattern.html

2) Enums:

This is a numeric value so less than operators could be an easy solution.