# Loops, If's, Case and other coding stuff...

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I had a thought while I was at work... perhaps we can create a resource for all of us and do it the easy way. There are many constructs that a lot of us struggle with and by that I mean that we can't get things put together solidly enough to even start to play with them and experiment. I hope that this thread will eventually become a sticky and be a resource for the syntax of various constructs, in each different kind of situation.One rule: To ask for something, you have to offer something... a tip, an example syntax, an explanation of why something works the way it does... anything. Offer something you know and then ask for something you'd like someone else to explain. I'll kick it off:************************This isn't the one I was hoping to put here but testing revealed a syntax problem in my first choice that I haven't been able to fix yet...It's been mentioned to use macros in order to save yourself a lot of work when it comes to typing out repeated values such as

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Scott,I'll give my two cents, starting inverse:*****************************************************%((@c:WaypointAirportFrequenciesNumber) s2 0 !=)--> this line retrieves max WaypointAirportFrequenciesNumber and tests whether it is greater than 0, if true (1) then the entire code that follows will execute:%{if}%(0 sp1) ---> The initial "index" for WaypointAirportFrequenciesNumber%{loop} ---> Start the loop until (#) condition %(l1 (>@c:WaypointAirportCurrentFrequency)) --> assigns the index value to WaypointAirportCurrentFrequency, so to be able to retrieve the related var values . Index starts at 0 and raises until it is equal to l2 (max WaypointAirportFrequenciesNumber)%((@g:listCurrent) l1 == (@g:listItems) 1 ==and)%{if}{blnk}%{end}%((@c:WaypointAirportFrequencyName))%!s!{nr}t%((@c:WaypointAirportFrequencyLimit))%{case}%{:1}{dpl=RX}%{:2}{dpl=TX}%{end}t%((@c:WaypointAirportFrequencyValue,Mhz) 100 * near d 100div)%!03d!.{fnt1}%(100 %)%!02d!{fnt}n--->Does a bunch of conditional prints %(l1 ++ s1 l2 <) --->(#) condition, increments the "index" by one and compares it with l2 (max WaypointAirportFrequenciesNumber)%{next}---> loops while the stack's last value is true (1) otherwise it resumes executing the rest of the code%{end}***********************************************************Hope that I didn't miss anything :-). Notice that this kind of {loop/next} construction applies only to structures, and is very useful just when dealing with conditional string outputs. But if you need to work with pure data, IMHO it is way better to use stack operators directly. Regarding your first "big" macro:Suppose you write a 100-line macro named "Macro1" and then call it once in a gauge like @Macro1(lvar,value,etc)It would be exactly the same to put the entire group of lines whithin the and not use the macro.Remember that macros are recognized only by code inside the gauge where they are definend, and are not visible by code inside other gauges. However, you can "repeat" them into as many gauges as you wish, even using the same name.And again, in your "st"{:1}{:2} macro example, what you are placing inside it is a whole set of string code, it will never return a stack value because it is precisely a string construction. So you can't do this "@st(nnn) (>LVarValue)" UNLESS a lonely and "isolated" value is left forgotten (or ex-profeso) in any of those %((var))% stack's portions.As already stated, my two (dollar would be ok for you?) cents ...Tom

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Tom,Thanks for the pointers. Let's see if I have this right:>%((MAX_ITERATIONS) s2 0 !=)>%{if}Places Max_Iterations into s2, compares s2 to 0. If false, run the code below.>%(0 sp1)Set 0 into sp1, this will be used to track the current iteration number, beginning with 0 and incrementing. >%{loop}Start the silly loop already... what are you waiting for?>%(l1 (>@c:WaypointAirportCurrentFrequency))I'm still fuzzy on the retrieval setup here... maybe it will come to me as I play with it some more. You said: >--> assigns the index value to>WaypointAirportCurrentFrequency, so to be able to retrieve the>related var values . Index starts at 0 and raises until it is>equal to l2 (max WaypointAirportFrequenciesNumber)I kind of got lost there... but that's more about retrieving the info than running the loop itself.>%((@g:listCurrent) l1 == (@g:listItems) 1 ==>and)%{if}{blnk}%{end}>%((@c:WaypointAirportFrequencyName))%!s!{nr}t>%((@c:WaypointAirportFrequencyLimit))%{case}%{:1}{dpl=RX}%{:2}{dpl=TX}%{end}t>%((@c:WaypointAirportFrequencyValue,Mhz) 100 * near d 100>div)%!03d!.{fnt1}%(100 %)%!02d!{fnt}>n>--->Does a bunch of conditional prints Like you said... it does stuff for each iteration.>%(l1 ++ s1 l2 <) Here's the baby... now that you did stuff, pull out l1 increment it and then shove it back into s1. Oh, and by the way, while you do that, check to make sure that it's less than the value we set into l2 at the start of all this. If it's still less even after being incremented:>%{next}Go around again. Otherwise, if it's equal to l2 then:>%{end}***************>Hope that I didn't miss anything :-). Notice that this kind of>{loop/next} construction applies only to structures,>and is very useful just when dealing with conditional string>outputs. But if you need to work with pure data, IMHO it is>way better to use stack operators directly. So this would or would not be the same sort of loop you would use for a timer?>Regarding your first "big" macro:>>Suppose you write a 100-line macro named "Macro1" and then>call it once in a gauge like >@Macro1(lvar,value,etc)>It would be exactly the same to put the entire group of lines>whithin the and not use the macro.Right, it's only when you have to call the same things many times over, not just one or two, that the macro begins to show any real benefit. Also, it only benefits the coder and his or her fingers, not the code... because the code is compiled with the full macro spelled out everywhere it gets called.>Remember that macros are recognized only by code inside the>gauge where they are definend, and are not visible by code>inside other gauges. However, you can "repeat" them into as>many gauges as you wish, even using the same name.True, but if the returned value of a macro can be placed into a numeric L:Var, then that L:Var could be seen and used by another gauge. The question is, how to get " @Macro (>L:Var, number) " to work... if it can be done.>And again, in your "st"{:1}{:2} macro example, what you are>placing inside it is a whole set of string code, it will never>return a stack value because it is precisely a string>construction. So you can't do this "@st(nnn) (>LVarValue)">UNLESS a lonely and "isolated" value is left forgotten (or>ex-profeso) in any of those %((var))% stack's portions.I know in the above I used strings, but what I'm uncertain of is would the numeric value of an A:Var be able to be called up by a macro and inserted into an L:Var, without the string typing?IE:Instead of using:%{:7}%((A:Autopilot heading lock dir, degrees))%!3d!Could we use:%{:7}%(A:Autopilot heading lock dir, degrees)%or:%{:7}(A:Autopilot heading lock dir, degrees)Or whatever syntax returns the 3 digit numeral that's in that variable.I think that I have seen such case statements used to set {img} and call macros... I wonder...1:00 AM here... work in the morning... to be continued...Tom, once again, thanks for clearing the original question up!Scott / Vorlin

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Scott,*****************************************>%((MAX_ITERATIONS) s2 0 !=)>%{if}Places Max_Iterations into s2, compares s2 to 0. If false,run the code below.******************It's the other way. If TRUE, then run the code below.As for REGISTERS, they are memory areas used to store values that come from the stack. Instead of saving a value into a LVar o GVar, you can save it straight into a register and be able to retrieve it whenever is needed across the entire gauge. They are reset to 0 at each FS cycle, and also when a "c" operator is placed on the stack. There are 50 available, from 0 to 49.To save a value into a register WITHOUT removing it from the stack, just precede its number with "s": 45 s0 ; (LVar,unit) s3If you need to save and remove the value, write "sp" before the number: 56.34 sp4 etcTo recall a register's value, write "l" before the number : l0, l30, etc................When working with gpsdll routines, it is necessary first to pass the parent function an index so the related child funtions will return their proper valuesFor example:7 (>@c:WaypointAirportCurrentFrequency)--> means you want to make the current frequency the one is situated on the 7th position.(@c:WaypointAirportFrequencyName) --> will put on the stack the name of the current airport frequency, which means the same as to be the name of the 7th airport frequency.......................******************So this would or would not be the same sort of loop you woulduse for a timer?***************I don't know what kind of timer you are speaking of.**************************True, but if the returned value of a macro can be placed intoa numeric L:Var, then that L:Var could be seen and used byanother gauge. The question is, how to get " @Macro(>L:Var, number) " to work... if it can be done.********************Think of macros as an extension of a very "huge" stack line.Suppose you have this simple one:34 5 + 38 + (>LVar1,unit)Now using a macro:38 + 34 5 + @My (>LVar1,unit)Translate this easy example to a complex macro struct and it will behave the same.Hope all these stuff was clear enough :-)Tom

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Scott / Vorlin,Did you ever try the code snippet you posted?

I have been doing a lot of graphics work over the past coupleof days and so haven't had time to try something like thispseudo-code:@2 1 == if{( @1 1 ==) if{ (A:Heading, number) (>L:CV1 ) } ( @1 2 ==) if{ (A:OBS1, number) (>L:CV1 ) } ( @1 3 ==) if{ (A:OBS2, number) (>L:CV1 ) } etc, etc..........}@2 2 == if{( @1 1 ==) if{ (A:Heading, number) (>L:CV2 ) } ( @1 2 ==) if{ (A:OBS1, number) (>L:CV2 ) } ( @1 3 ==) if{ (A:OBS2, number) (>L:CV2 ) } etc, etc..........}@2 3 == if{( @1 1 ==) if{ (A:Heading, number) (>L:CV3 ) } ( @1 2 ==) if{ (A:OBS1, number) (>L:CV3 ) } ( @1 3 ==) if{ (A:OBS2, number) (>L:CV3 ) } etc, etc..........}Then when I need to call any A:Var for the current value, allI'd need to do is @My(GaugeWanted, Var to put it in)Such a setup can allow all gauges to be written to work withL:CV(n) (CurrentVar#).
Did it work? If so that seems like an excellent way to use a macro for indexing to store a passed variable.R/TC2

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Hi,I quickly adapted Scott's example to show a way that works great when dealing with arrays of LVars :Suppose you have an array of 100 lvars and want to assign a value to the index 4, l0 @1 == if{ l3 l2 l1 0 4 l20 case (>L:CV@1,number) }And the calling code:(L:Index,enum) sp0 (* The index value, 4 in this case *)(A:Heading, number) sp1(A:OBS1, number) sp2(A:OBS2, number) sp3(L:HeadObs1Obs2,enum) sp20 (* A counter from 1 to 3 that tells whether Heading, OBS1 or OBS2 would be used *)@My(1) @My(2) @My(3) @My(4) ...@My(100) Only (L:CV4,number) will be updated.This is very simple, but you can handle multidimensional arrays as well with few lines of code. Tom

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Tom, brilliant as usual! Thanks for your reply - this looks like a dandy way to cotrol arrays of stored data. With some modification, the concept should be able to work well in the RMS gauge I am currently working on. Thank you!R/TC2

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Ok Tom, I've shown in my gauge that I can transfer an index into a macro and have it use that index for storing a value into that indexed variable.Here's the macro:@2 (>L:ch@1,enum)and the call is:@chst(211,11811).This works fine, and loads (L:ch211,enum) with the value 11811. However, when I tried to pass a defined L variable back, to either call up or store into an indexed variable, it doesn't seem to work. For example:The Macro:@2 (>L:ch@1,enum)The call:@chst((L:achan,enum),11811)This doesn't work, and results in an "undefined" LVar.Similarly,Macro:(L:ch@1,enum) (>L:afreq,enum) (>L:scratch,enum)and the call:@acfreq((L:achan,enum))does not load the defined value of (L:achan,enum) into either of the (L:afreq,enum) or (L:scratch,enum) variables.I read in another thread that (L:Var,unit) can be passed in a call as a valid argument to a macro, but it doesn't look like a LVar argument can be used within the structure of another LVar in a macro. Can you confirm that this is the case? If so, might you know of any other way to send an indexed value in a macro call to a LVar within a macro? I tried it just now using a register:The Macro:(L:ch@1,enum) (>L:afreq,enum) (>L:scratch,enum)The call:(L:achan,enum) sp0 @acfreq(l0)failed to send the value of (L:achan,enum), stored in register 0, to call up (L:ch[l0's value, equal to (L:achan,enum)], enum). Instead, it created a LVar with the name "chl0". So that method didn't work either.R/TC2

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To continue...in another thread, Tom provided an outstanding answer to the following question:

>would you be kind enough to explain the significance of the @>character, that is, what does @1, @4, etc. achieve?Sure The @ character symbolizes a paramater passed to a macro, and the number to the right is the order in the sequence of the total parameters passed, each of them separated by comma.For example @1 2 * @2 + @3 +Then, calling @Macro1(10,20,30) will pass 10 to @1, 20 to @2 and 30 to @3.Now, what kind of values can receive a parameter "@" ?1) literal stack values : @Macro1(10)2) A,L,G var values : @Macro1((A:Indicated Altitude, feet)) 3) Register values : @Macro1(l0,l1,l2)4) Part of var names : @Macro1(A:Indicated Altitude) or @Macro1(feet)5) Macro names : @Macro1(@NestedMacro)6) Resultant stack values : @Macro1(5 2 *,4 3 +) will pass 10 and 77) "Popped" stack values : 20 @Macro1(10) will pass 10 to @1 AND 20 to @2 if that exists in the called This is an essential description, hope it was clear enough
However, in experimenting tonight, I've found the following clarification to share with you all which may help your ubderstanding of the way macro arguments work:If the argument sent is embedded within another variable's name definition, the argument sent is a literal string, not the value that the

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Well, I see you've figured out the whole thing by yourself :-)I've stated in a bunch of earlier threads that macros are NOT functions but literal replacement of text. Therefore, anything you pass as an argument will be "inserted" in the interpreted code "as is" and, as you cleverly found, will give different results on each case, depending on its final meaning.You may notice that I wrote @My(1) @My(2) ..etc to reference a call to each "indexed" Lvar, precisely because of the limitation stated by you. Surely would be way better to have the possibility to write @My((L:Index,enum)) and pass the value of the var and not a simple text (which would be the case if a macro could work like a real function);I praised for that to be supported in FSX XML, maybe using the current "index" of an lVar( currently (L:Var1:1,number) and (L:Var1:2,number) are recognized as different) but MS didn't make any change on this, probably because it is technically impossible with the current schemas.Despite of this, it is not so tough to write a string of 100 macros if you use short names and pass only one argument :-)Also it is rather easy to handle multidim arrays, using a bunch of (a bit complex) macros...Tom

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Hmmm, Tom, the following gives me cause to think:

7) "Popped" stack values : 20 @Macro1(10) will pass10 to @1 AND 20 to @2 if that exists in the called
What if I tried something like:Assumptions: (L:ch231,enum) is an LVar where I want to store a freq;(L:achan,enum) has a value of 231 assigned elsewhere;(L:myfreq,enum) has a value of 12345 assigned elsewhereThe Macro:@1 (>L:ch@2,enum)The call:(L:achan,enum) @afreq((L:myfreq,enum))Wouldn't this then "pop" the value of (L:achan,enum) off the stack (not the string representation of "(L:achan,enum)" )and pass the popped value of 231 to the macro so I would end up with:12345 (>L:ch231,enum) when the macro executes??Might this work? I'll try it tonight...R/TC2

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Hello ScottI was looking at your macro post today and wonder:>This way, if you want the airspeed to be displayed digitallythen all you need to do is use @st(8).Like this? @st(8)Just to make sureRoelof

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TC2,<>No it won't, unfortunately. The post you quoted <<20 @Macro1(10)>> will work if the macro contains @1 and @2 as stack values and not being a string part of a variable. Funny to see you are doing the same tests I did almost a year ago...the best way to learn XML is never give up testing :-)Tom

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Of course, you are right, Tom :) Thought I'd give it a try anyway, but it was close! Here's the results...The code:@1 (>L:ch@2,enum)%((L:ch456, enum) 100 /)%!5.2f!%((L:ch123, enum) 100 /)%!5.2f!13600 (>L:myfreq,enum) 123 (>L:mychan,enum) (L:mychan,enum) @test((L:myfreq,enum))11800 (>L:myfreq2,enum) 456 (>L:mychan2,enum) (L:mychan2,enum) @test((L:myfreq2,enum))And the results:1) The gauge worked (i.e. syntax was OK).2) Using XML_ID, the L:Vars ended up with the following values: (L:myfreq,enum)=13600 (L:myfreq2,enum)=11800 (L:mychan,enum)=123 (L:mychan2,enum)=456 as expected. When the macro ran: a. Macro called first time: @1=13600, @2 pulled and popped from the stack with a value of 123 b. Macro created (L:ch123,enum)=00000 c. Macro stored (L:myfreq,enum)=13600 into (L:ch,enum) (not into (L:ch123,enum) as hoped. d. Macro called the second time: @1=11800, @2 pulled and popped from the stack with a value of 456 e. Macro created (L:ch123,enum)=00000 f. Macro stored (L:myfreq2,enum)=11800 into (L:ch,enum) (not into (L:ch456,enum) as hoped.3) Final result: (L:ch123,enum)=00000 (L:ch456,enum)=00000 (L:ch,enum)=11800 (the last macro operation performed)So...close! But no cigar! It DID pull the VALUE from the stack (not the "string representation") and passed it to the macro as a valid argument, but it did not allow that argument to be concatenated with "ch" to identify the proper variable to place the @1 argument into. Looks like I have once again proven that the Master (Tom Taquilo) was right on the money, and that we will not be able to use macros for array functionality.So I am stuck with lots of "if{" statements to do a chore that, with a simple variable array call, could have been one short line. Oh well! Back to coding if{ statements :)R/TC2(and props to the Master...Tom T.)

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TC2,Nice try! :-), good for you. There's no other way to master XML code than doing an "endless loop" of tests :D""So I am stuck with lots of "if{" statements to do achore that, with a simple variable array call, could have beenone short line. Oh well! Back to coding if{ statements :)""Why is that? You only need a few if{ statements for simple arrays (see my former example in this thread)Now, as you're a die hard tester :-), take a look at this macro@1(1) @1(2) @1(3) @1(4) @1(5) @1(6) @1(7) @1(8)@1(9) @1(10) @1(11) @1(12) @1(13) @1(14) @1(15) @1(16)@1(17) @1(18) @1(19) @1(20) @1(21) @1(22) @1(23) @1(24)@1(25) @1(26) @1(27) @1(28) @1(29) @1(30) @1(31) @1(32)@1(33) @1(34) @1(35) @1(36) @1(37) @1(38) @1(39) @1(40)It can't be said that there is a lot of code here. But with this core macro you can add, retrieve, delete, insert, invert, values within an array of 40 LVars. Maybe you'd want to figure out a way on how to use it in a real example...Regards,TomPD: I'm not a Master, just a tough tester *:-*

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Thanks Tom. Based on one of your other clever suggestions for loops, I tried to use one in a macro, however, it isn't providing the expected results - it isn't looping. The gauge its used in appears, so there are no syntax errors that the parser can't understand, but the macro isn't performing. Note that my channel indices are not contiguous, but case allows them to become numerically contiguous. Here's the code: %((L:field,enum) 10 / int 1 + (L:mempgnr,enum) 1 - 5 * + sp2)%(20 sp1) %{loop} %(l1)%{case}%{:0}%{:1}%(0 (>L:ch211,enum))%{:2}%((L:ch211,enum) (>L:ch221,enum))%{:3}%((L:ch221,enum) (>L:ch231,enum))%{:4}%((L:ch231,enum) (>L:ch241,enum))%{:5}%((L:ch241,enum) (>L:ch251,enum))%{:6}%((L:ch251,enum) (>L:ch212,enum))%{:7}%((L:ch212,enum) (>L:ch222,enum))%{:8}%((L:ch222,enum) (>L:ch232,enum))%{:9}%((L:ch232,enum) (>L:ch242,enum))%{:10}%((L:ch242,enum) (>L:ch252,enum))%{:11}%((L:ch252,enum) (>L:ch213,enum))%{:12}%((L:ch213,enum) (>L:ch223,enum))%{:13}%((L:ch223,enum) (>L:ch233,enum))%{:14}%((L:ch233,enum) (>L:ch243,enum))%{:15}%((L:ch243,enum) (>L:ch253,enum))%{:16}%((L:ch253,enum) (>L:ch214,enum))%{:17}%((L:ch214,enum) (>L:ch224,enum))%{:18}%((L:ch224,enum) (>L:ch234,enum))%{:19}%((L:ch234,enum) (>L:ch244,enum))%{:20}%((L:ch244,enum) (>L:ch254,enum)) %(l1 -- s1 l2 >)%{next}%{end}(L:field,enum) carries a value of 10,20,30,40 or 50 and a single value is already present when the macro is called.(L:mempgnr,enum) carries a value of 1,2,3 or 4 and a single value is already present when the macro is called.The values of (L:field,enum) and (L:mempgnr,enum) determine the stop index (which carries a value between 1 and 20, inclusive). For example, if (L:field,enum) = 30 and (L:mempgnr,enum)=1, then the value stored in register 2 should be:Calculations in RPN:30 10 / => 3 3 int => 33 1 + => 41 1 - => 00 5 * => 04 0 + => 44 => sp2The loop should start by executing case 20, then 19, 18, ...4, and then stop.The call is @insert and no arguments are passed.The call results in no evidence that the macro is working - no channel values change when the call is made. I can't read the registers, so I've no idea where the loop is stopping.I also tried it by making the {case} part into a separate macro and calling it from within the "Insert" macro with an argument of (l1). That didn't work either - same symptoms.Any idea what might be causing this to not loop? or other suggestions?R/ and many thanks for all your help so far,TC2(Tom Stevenson, Montclair Virginia USA)

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Tom,I think the correct syntax should be:{loop}value{case}{:1}...{:20}{end}conditional{next}No need to use {:0} if you won't have a value here; works different from stack's "case" structure.And I assume you are calling the macro within a tag?- like @insert - because if called from inside a it won't work. Tom

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Tom, as usual, you are correct. Despite the syntax change, the loop failed to work properly when called from a click statement. So I went back to the drawing board, and, using macros and a slightly different arrangement of my array (inspired by your 40 item single array), made it work! Here's the code for those interested:(L:field,enum) 10 / int (L:mempgnr,enum) 1 - 5 * + (>L:chnnbr,enum) (L:chnnbr,enum)@1 20 < if{ (L:ch244,enum) (>L:ch254,enum) }@1 19 < if{ (L:ch234,enum) (>L:ch244,enum) }@1 18 < if{ (L:ch224,enum) (>L:ch234,enum) }@1 17 < if{ (L:ch214,enum) (>L:ch224,enum) }@1 16 < if{ (L:ch253,enum) (>L:ch214,enum) }@1 15 < if{ (L:ch243,enum) (>L:ch253,enum) }@1 14 < if{ (L:ch233,enum) (>L:ch243,enum) }@1 13 < if{ (L:ch223,enum) (>L:ch233,enum) }@1 12 < if{ (L:ch213,enum) (>L:ch223,enum) }@1 11 < if{ (L:ch252,enum) (>L:ch213,enum) }@1 10 < if{ (L:ch242,enum) (>L:ch252,enum) }@1 9 < if{ (L:ch232,enum) (>L:ch242,enum) }@1 8 < if{ (L:ch222,enum) (>L:ch232,enum) }@1 7 < if{ (L:ch212,enum) (>L:ch222,enum) }@1 6 < if{ (L:ch251,enum) (>L:ch212,enum) }@1 5 < if{ (L:ch241,enum) (>L:ch251,enum) }@1 4 < if{ (L:ch231,enum) (>L:ch241,enum) }@1 3 < if{ (L:ch221,enum) (>L:ch231,enum) }@1 2 < if{ (L:ch211,enum) (>L:ch221,enum) }@1 1 < if{ 0 (>L:ch211,enum) }@delete(@channum) (L:scratch,enum) (>L:ch254,enum) 0 (>L:insflg,bool)The calls are very simple (from click statements):If I want to insert a 0 as a value into a channel in the array of memory channels, shifting all channel values down 1 space, I set a flag, then I store the value of channel 254 (the 20th channel in the array) in a scratch variable in case I need to undo it. Then I call the insert macro using macro @channum as the argument. Channum sets the index value to stop at (start at in the case of a @delete call) the active channel (which is known already when the call is made). So the full call is:1 (>L:insflg,bool) (L:ch254,enum) (>L:scratch,enum) @delete(@channum)If I want to undo the insert, all I have to do is call:@undoins and voila! everything is restored back to where it was before the insert call!Delete and UndoDel work in the same fashion.Tom, thanks a million for all the help on this! On to coding the next subsection!Many warm regards,TC2

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