Best practice with do loops

Questions
1

Hey, I have a beginners question:
How can I write this shorter. I’m pretty sure there is a way, but can’t find it in the help:

(
i = 0;
8.do{
	i=i+1;
	('Execute'++i).postln;
}
)
2

Perhaps one of the below?

8.do({ arg i; (i + 1).postln })
1.to(8).do({ arg i; i.postln })
(1 … 8).do({ arg i; i.postln })

Ps. The “Look Up Implementations…” entries in the “Language” menu of the editor can be a nice way of finding things…

3 
8.do{|i|('Execute'++i).postln};
4

thanks! I didn’t know about the “Loop Up Implementation”. That helps

5

That’s exactly what I was searching for. Thanks! So the argument counts upwards automatically!

6 
8.do{ |i| (\Execute ++ (i + 1)).postln }

(1..8).do{ |i| (\Execute ++ i).postln }