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> Main > Using macro 2+3/2-3*2 replace 2,3/2,3*2 |
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Roman 15 Jul 2013, 10:29
That no one knows how to do it?
Wow !  |
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15 Jul 2013, 10:29 |
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bitRAKE 15 Jul 2013, 11:20
l_inc posted something that is very close to your desired goal. Instead of displaying the values, they would need to be compared against your list of symbols, and alternate action taken in that case.
Code: struc asComputationDestination { macro . [\.] \{ common match ==\\.,\. \\{ irps \\\.,\\. \\\{ __out equ \\\. match +,\\\. \\\\{ restore __out __out equ , \\\\} match -,\\\. \\\\{ restore __out __out equ , \\\\} match \\\\.,__out \\\\{ display \\\\`\\\\. \\\\} restore __out \\\} \\} \} } x asComputationDestination x = 20*10/20+1 display 13,10 x = 2+3/2-3*2 display 13,10 x = 2+3/2-3*2-4+3/2-1 
Despite this code looking gnarly, __out is the only part I've added. It either remains the current symbol, or is replaced by a comma. _________________ ¯\(°_o)/¯ “languages are not safe - uses can be” Bjarne Stroustrup |
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15 Jul 2013, 11:20 |
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Roman 15 Jul 2013, 11:47
bitRAKE
Fantastic ! Thanks. How you write this macro? How then can you understand that macro? |
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15 Jul 2013, 11:47 |
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Roman 15 Jul 2013, 11:59
bitRAKE
You macro cool ! But you macro display. I need get (for example NewResult) NewResult EQU 2,3/2,3*2 and then i use NewResult in next steps of code. I mean thats. |
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15 Jul 2013, 11:59 |
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bitRAKE 15 Jul 2013, 13:03
Code: struc Parse0 [A] { common . equ irps B,A \{ __out equ B match +,B \\{ restore __out __out equ , \\} match -,B \\{ restore __out __out equ , \\} match C,__out \\{ match D,. \\\{ restore . . equ D\\\#C \\\} match ,. \\\{ restore . . equ C \\\} \\} restore __out \} } y Parse0 2+3/2-3*2-4+3/2-1 ; y EQU 2,3/2,3*2,4,3/2,1 ; can even verify with previous macro x = y The FASM manual is pretty good - read it a couple times; keep it open as a reference. Use the \FASM\TOOLS\* to see what happens behind the scenes. Little by little it will become clear. Last edited by bitRAKE on 15 Jul 2013, 13:45; edited 2 times in total |
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15 Jul 2013, 13:03 |
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Roman 15 Jul 2013, 13:13
bitRAKE Thanks !
Sorry. What is FASM\TOOLS ? I never used FASM\TOOLS |
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15 Jul 2013, 13:13 |
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bitRAKE 15 Jul 2013, 13:36
I believe it's in the download.
LISTING PREPSRC SYMBOLS Read here: http://board.flatassembler.net/topic.php?t=9792 |
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15 Jul 2013, 13:36 |
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l_inc 15 Jul 2013, 14:34
bitRAKE
Quote: You help me understand because I try to help wherever I can. Read, read, read... Often I fail, and I try again. Then one day I see something I didn't see before. Consider two things regarding your macro: 1) You're using the match directive extensively, even though it's not needed. Remember, that match and equ perform symbol expansion. Thus in the following case I would expect y to have the value var1*2, not var5*2 . Code: define var5 50 define var4 var5 define var3 var4 define var2 var3 define var1 var2 define exp1 var1*2 match x,exp1 { y Parse0 x } 2) The concatenation operator is incorrectly used. No symbol concatenation is needed. This way the following code should give y the value "2 x" and not the value "2x": Code: define exp 2 x match x,exp { y Parse0 x } _________________ Faith is a superposition of knowledge and fallacy |
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15 Jul 2013, 14:34 |
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bitRAKE 15 Jul 2013, 15:06
Feel free to provide a more general solution.
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15 Jul 2013, 15:06 |
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l_inc 15 Jul 2013, 15:22
bitRAKE
This one seems to be quite general: Code: ;Allows to substitute one symbol in a list with another symbol ;usage: expression equsubs var2,var1,sqrt(var1)/var1*2+3 struc equsubs symnew, symold*, [arg] { common restore . . equ res2 equ irps sym,arg \{ \forward define res1 sym match =symold, res1 \\{ restore res1 define res1 symnew \\} \forward res2 equ res1 res2 restore res1 \common restore . . equ res2 \forward restore res2 \} restore res2 } _________________ Faith is a superposition of knowledge and fallacy |
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15 Jul 2013, 15:22 |
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bitRAKE 15 Jul 2013, 16:00
Nice execution. Don't understand why the prior value of (.) should be lost though, but that's just me.
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15 Jul 2013, 16:00 |
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l_inc 15 Jul 2013, 17:22
bitRAKE
Quote: Don't understand why the prior value of (.) should be lost though That's for convenience. I don't like macro or constant stacks to arbitrarily accumulate. Kinda principle of least side effect. Thus every time a logical scope of a constant (or a macro) is closed it should be restored (or purged). For that reason it's easier to include restore into the constant transformation routines than to explicitly invoke restore each time a constant needs to be transformed: Code: ;Allows to assign new value to a symbol without congesting it's stack ;usage: mysym equ 1 ; mysym reequ mysym,2 struc reequ [val] { common tmp equ val restore . . equ tmp restore tmp } ;Allows calculations with preprocessor ;usage: result equcalc 15/2 mod result struc equcalc expr { rept 1 res:expr \{ restore . . equ res \} } ;Finds the maximum value ;usage: max equmax 10,-15 struc equmax val1*, val2* { rept 1 diff: ((val1)-(val2)) \{ restore . . equ val1 match - any,diff \\{ restore . . equ val2 \\} \} } ;Finds the minimum value ;usage: max equmin 10,-15 struc equmin val1*, val2* { rept 1 diff: ((val1)-(val2)) \{ restore . . equ val2 match - any,diff \\{ restore . . equ val1 \\} \} } ;Allows to replicate the specified symbol a number of times ;with an optional delimiter ;usage: replication equdup 'hello',10,<#',',> ; display replication struc equdup val*, count*, delim { tmp equ tmp equmax 1,count restore . . equ rept tmp-1 \{ . reequ . val delim \} rept 1 i:count \{ match =i,tmp \\{ . reequ . val \\} \} restore tmp } ;Allows to concatenate values of the arguments ;usage: concatenation equcat x,y,z ; concatenation: struc equcat [arg] { common tmp equ forward match val1,arg \{ match val0,tmp \\{ restore tmp define tmp val0\\#val1 \\} match ,tmp \\{ restore tmp define tmp val1 \\} \} common restore . . equ tmp restore tmp } ;and so on ... This approach also corresponds to the generally accepted requirement to define variable before using: Code: myvar1 equ 3 ;define myvar1 myvar2 equ 2 ;define myvar2 myvar1 equcalc myvar1*4 myvar2 equcalc myvar2*5 tmpvar equ myvar1 ;define tmpvar myvar1 equmin myvar1,myvar2 myvar2 equmax tmpvar,myvar2 lbl equ ;define lbl with dummy value lbl equdup x,myvar1,<,_,> match x,lbl { lbl equcat x,_,myvar2 } lbl: restore myvar1,myvar2,tmpvar,lbl ;destroy all constants if defined x_x_x_x_x_x_x_x_x_x_12 display 'We''ve just defined something completely useless',13,10 end if _________________ Faith is a superposition of knowledge and fallacy |
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15 Jul 2013, 17:22 |
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Roman 17 Jul 2013, 03:44
How to make a counter?
Example. We have V1 fix 2+1/2*3 and after using macro x Parse V1 we have x = 2,1/2*3 And need have count = 6. Number 6 is pointer to * befor 3. If we do restore count we get count= 4 (Number 4 is pointer to / befor 2) My code not work. Quote:
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17 Jul 2013, 03:44 |
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