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Question about gdscript: Which is more efficient? (sopuli.xyz)

submitted 1 year ago* (last edited 1 year ago) by [email protected] to c/[email protected]

21 comments fedilink hide all child comments

I’m just curious about which is the most efficient way of doing this kind of node enumiration:

for i in something():
    o=[var1,var2,var3,varN][i]
    o.new()
    o.do_something_based_on_number_of_loops()
    add_child(o)

for i in something():
    match i:
        0:
            o=var1
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)
        1:
            o=var2
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)
        2:
            o=var3
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)
        N-1:
            o=varN
            o.new()
            o.do_something_based_on_number_of_loops()
            add_child(o)

var items = [var1,var2,var3,varN]
for i in something():
    o=items[i]
    o.new()
    o.do_something_based_on_number_of_loops()
    add_child(o)

Or is there a more efficient way of doing it?

Edit: Sorry if that wasn't clear. Is it better to constantly get something from an "unstored list", store the list in a variable, or not use a list and use a match statement instead? Do they have any advantages/disadvantages that make them better in certain situations?

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[–] [email protected] 1 points 1 year ago (2 children)

It's a poor name choice then, because it actually says less about what it's doing than the main function does.

Besides, what is the point of "looking further" just to stop at another function name? Wouldn't looking further imply the need to review the implementation?

[–] tabular 1 points 1 year ago

In practice it turns out the method to make just the first element visible was redundant anyway. It would be made visible during the setup function that all elements call.