OpenEuphoria

2. Re: routine_id

• Posted by DerekParnell (admin) Aug 25, 2008
• 859 views

Ask yourself, in that program, where is the call statement? Your program isn't calling it. It's the Win32lib library that is calling your routine, and its call statement is before your routine's declaration.

However, here is a simpler example.

integer r_MyProc 
 
call_proc(r_MyProc, {})  --<<<< Here is the call. 
 
procedure MyProc() --<<<< Here is the routine. 
  puts(1, "MyProc called\n") 
end procedure 
r_MyProc = routine_id("MyProc") 

-- V4.0 -- 
MyProc()  --<<<< Here is the call. 
 
procedure MyProc() --<<<< Here is the routine. 
  puts(1, "MyProc called\n") 
end procedure 

4. Re: routine_id

• Posted by mattlewis (admin) Aug 25, 2008
• 847 views

integer r_MyProc 
 
call_proc(r_MyProc, {})  --<<<< Here is the call. 
 
procedure MyProc() --<<<< Here is the routine. 
  puts(1, "MyProc called\n") 
end procedure 
r_MyProc = routine_id("MyProc") 

Which is actually broken, since you used r_MyProc before it was assigned:

integer r_MyProc 
procedure main() 
  call_proc(r_MyProc, {})  --<<<< Here is the call. 
end procedure 
 
procedure MyProc() --<<<< Here is the routine. 
  puts(1, "MyProc called\n") 
end procedure 
r_MyProc = routine_id("MyProc") 
 
main() 

6. Re: routine_id

• Posted by DonCole Aug 26, 2008
• 824 views

integer r_MyProc = routine_id("MyProc") 

Thank you Derek,

I see you must use call_proc() first.

I didn't know that.

And gShingles,

I type my post in TextPad first (because it has SpellCheck ; this forum does not.

In all fairness nether did the old forum.)

I copy then paste from TextPad to the forum.

This where I believe <eucode> and </eucode> get lost.

Don Cole A Bug is an un-documentedfeature. A Feature is a documented Bug.

8. Re: routine_id

• Posted by DonCole Aug 26, 2008
• 809 views

Sorry Derek,

When I run your program I get variable r_MyProc has not been assigned a value .

Don Cole A Bug is an un-documentedfeature. A Feature is a documented Bug.

10. Re: routine_id

• Posted by DonCole Aug 26, 2008
• 801 views

integer r_MyProc  
procedure main()  
  call_proc(r_MyProc, {})  --<<<< Here is the call.  
end procedure  
  
procedure MyProc() --<<<< Here is the routine.  
  puts(1, "MyProc called\n")  
end procedure  
r_MyProc = routine_id("MyProc")  
  
main() 

Yes I see that now.

But it looks like all this is being called from main() at the bottom of the code. I have to study this awhile.

Thanks again Derek.

Don Cole A Bug is an un-documented feature. A Feature is a documented Bug.

12. Re: routine_id

• Posted by mattlewis (admin) Aug 26, 2008
• 814 views

Actually, you don't even need 'declared' or anything like that in 4.0:

MyProc()  --<<<< Here is the call.   
   
procedure MyProc() --<<<< Here is the routine.   
  puts(1, "MyProc called\n")   
end procedure   

Matt

14. Re: routine_id

• Posted by mattlewis (admin) Aug 26, 2008
• 808 views

Yes.

Matt

16. Re: routine_id

• Posted by mattlewis (admin) Aug 26, 2008
• 799 views

How does ver 4.0 know what the parameters are ? 
 
Are you doing a 2 pass parse ? 
 
I noticed that the startup parse seems to be quite slow on my 900MHz machine. 
 

It doesn't. What it does is leave some extra space in case of extra default parameters, and also remembers the fact that there was something that looked like a forward reference. It then checks (at the end of each file) to see if it can resolve any of the references that weren't previously resolved.

At the very end, if there are still any outstanding calls that have yet to be resolved, they are listed as an error.

So each forward reference takes a little bit more room in the IL than a normal call would, and there's a slight performance difference, since there's an extra goto in the IL to jump from the end of the actual call to the beginning of the next instruction. This is pretty minor (and you'd probably have to really work at it to tell the difference). It is definitely faster than a call_proc/func though.

Matt

18. Re: routine_id

• Posted by euphoric (admin) Aug 26, 2008
• 807 views

I noticed that the startup parse seems to be quite slow on my 900MHz machine. 

I thought last week the parse time was slower. That was only my "perception," and nothing based on hard numbers. What is the speed of the parser pre-FR compared to now? (Just curious and to make sure I'm perceiving things accurately. )

20. Re: routine_id

• Posted by euphoric (admin) Aug 26, 2008
• 794 views
• Last edited Aug 27, 2008

Uh oh. That sounds bad.

Jeremy, you remember when I brought that up in chat last week, about the parser being slower? Maybe it was a different change to the interpreter (nothing to do with FR). I shoulda posted it to the dev list. Drat!

22. Re: routine_id

• Posted by mattlewis (admin) Aug 27, 2008
• 854 views

No, nothing has changed there. There are still times when they are the best solution.

Matt