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> Macroinstructions > help with some macros |
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revolution 30 Jun 2017, 10:43
Okay, your macro is tricky because equ is a textual replacement and things get confusing quickly.
Rather than try to fix it manually here I would like to suggest you assemble the "PREPSRC.ASM" file (which can be found the TOOLS folder in the fasm download). Run that over your source code and examine the output to see all of the things that happened in the preprocessor stage during assembly. In that you will see all of the textual replacements that happen. |
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30 Jun 2017, 10:43 |
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username 30 Jun 2017, 12:19
ok, i remove the 'win32ax.inc' line, then i run the prepsrc over file.fas and got this output:
Code: format PE GUI 4.0 entry start ;eaxul equ eax ;ebxul equ ebx ;ecxul equ ecx ;edxul equ edx ;ediul equ edi ;esiul equ esi ;ebpul equ ebp ;macro ext instr ;{ ; macro instr op1,op2,op3 ; \{ ; instr op1,op2 ; if op3 eq ; ; else ; if(op3 mod 2)=0 ; exchange eaxul,ebxul ; else ; exchange ecxul,edxul ; end if ; end if ; \} ;} ;macro show ;{ ; display eax=eaxul ebx=ebxul ecx=ecxul edx=edxul edi=ediul esi=esiul ebp=ebpul ;} ;macro exchange var1,var2 ;{ ; local tmp1,tmp2,tmp3 ; ; tmp1 equ var1 ; tmp2 equ var2 ; ; xchg var1,var2 ; ; tmp3 equ tmp1 ; tmp1 equ tmp2 ; tmp2 equ tmp3 ; ; var1 equ tmp1 ; var2 equ tmp2 ;} section '.text' code readable executable start: xor eax,eax ;exchange eaxul,edxul ; tmp1?0 tmp2?1 tmp3?2 ;tmp1?0 equ eax ;tmp2?1 equ edx xchg eax,edx ;tmp3?2 equ eax ;tmp1?0 equ edx ;tmp2?1 equ eax ;eaxul equ edx ;edxul equ eax ;exchange ecxul,ediul ; tmp1?3 tmp2?4 tmp3?5 ;tmp1?3 equ ecx ;tmp2?4 equ edi xchg ecx,edi ;tmp3?5 equ ecx ;tmp1?3 equ edi ;tmp2?4 equ ecx ;ecxul equ edi ;ediul equ ecx ;exchange ecxul,eaxul ; tmp1?6 tmp2?7 tmp3?8 ;tmp1?6 equ edi ;tmp2?7 equ edx xchg edi,edx ;tmp3?8 equ edi ;tmp1?6 equ edx ;tmp2?7 equ edi ;ecxul equ edx ;eaxul equ edi ;exchange esiul,ebxul ; tmp1?9 tmp2?A tmp3?B ;tmp1?9 equ esi ;tmp2?A equ ebx xchg esi,ebx ;tmp3?B equ esi ;tmp1?9 equ ebx ;tmp2?A equ esi ;esiul equ ebx ;ebxul equ esi ;exchange ebxul,eaxul ; tmp1?C tmp2?D tmp3?E ;tmp1?C equ esi ;tmp2?D equ edi xchg esi,edi ;tmp3?E equ esi ;tmp1?C equ edi ;tmp2?D equ esi ;ebxul equ edi ;eaxul equ esi ;exchange ebpul,ediul ; tmp1?F tmp2?G tmp3?H ;tmp1?F equ ebp ;tmp2?G equ ecx xchg ebp,ecx ;tmp3?H equ ebp ;tmp1?F equ ecx ;tmp2?G equ ebp ;ebpul equ ecx ;ediul equ ebp ;exchange esiul,eaxul ; tmp1?I tmp2?J tmp3?K ;tmp1?I equ ebx ;tmp2?J equ esi xchg ebx,esi ;tmp3?K equ ebx ;tmp1?I equ esi ;tmp2?J equ ebx ;esiul equ esi ;eaxul equ ebx ;exchange edxul,ediul ; tmp1?L tmp2?M tmp3?N ;tmp1?L equ eax ;tmp2?M equ ebp xchg eax,ebp ;tmp3?N equ eax ;tmp1?L equ ebp ;tmp2?M equ eax ;edxul equ ebp ;ediul equ eax inc ebx ;exchange edxul,ecxul ; tmp1?O tmp2?P tmp3?Q ;tmp1?O equ ebp ;tmp2?P equ edx xchg ebp,edx ;tmp3?Q equ ebp ;tmp1?O equ edx ;tmp2?P equ ebp ;edxul equ edx ;ecxul equ ebp add ebx,2 ;exchange eaxul,ebpul ; tmp1?R tmp2?S tmp3?T ;tmp1?R equ ebx ;tmp2?S equ ecx xchg ebx,ecx ;tmp3?T equ ebx ;tmp1?R equ ecx ;tmp2?S equ ebx ;eaxul equ ecx ;ebpul equ ebx add ecx,2 !!!!!!!!!!!!!!UNTIL HERE EVERYTHING IS OK!!!!!!!!!!!!!!!!!!!! ;ext sub ;macro sub op1,op2,op3 ;{ ; sub op1,op2 ; if op3 eq ; else ; if(op3 mod 2)=0 ; exchange eaxul,ebxul ; else ; exchange ecxul,edxul ; end if ; end if ;} ;sub eaxul,1 sub ecx,1 if eq else if(mod 2)=0 ;exchange eaxul,ebxul ; tmp1?U tmp2?V tmp3?W ;tmp1?U equ ecx ;tmp2?V equ edi xchg ecx,edi !!!!!!!!!!!!this don't show in the exe, but eaxul gets edi!!!!!!!!!!!!!!!!!! ;tmp3?W equ ecx ;tmp1?U equ edi ;tmp2?V equ ecx ;eaxul equ edi ;ebxul equ ecx else ;exchange ecxul,edxul ; tmp1?X tmp2?Y tmp3?Z ;tmp1?X equ ebp ;tmp2?Y equ edx xchg ebp,edx ;tmp3?Z equ ebp ;tmp1?X equ edx ;tmp2?Y equ ebp ;ecxul equ edx ;edxul equ ebp end if end if ;sub eaxul,2 sub edi,2 if eq else if(mod 2)=0 ;exchange eaxul,ebxul ; tmp1?a tmp2?b tmp3?c ;tmp1?a equ edi ;tmp2?b equ ecx xchg edi,ecx ;tmp3?c equ edi ;tmp1?a equ ecx ;tmp2?b equ edi ;eaxul equ ecx ;ebxul equ edi else ;exchange ecxul,edxul ; tmp1?d tmp2?e tmp3?f ;tmp1?d equ edx ;tmp2?e equ ebp xchg edx,ebp ;tmp3?f equ edx ;tmp1?d equ ebp ;tmp2?e equ edx ;ecxul equ ebp ;edxul equ edx end if end if retn it seems that the preprocessor processes ALL the equ instructions first disregarding if they are inside some preprocessing condition, then checks the conditions to output the asm instructions? any workaround to this? |
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30 Jun 2017, 12:19 |
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revolution 30 Jun 2017, 12:26
The thing to realise is the separation of fasm's layers. The assembler stage will process the if, else and end if after the preprocessor has finished.
The "workaround" is to use preprocessor conditional structures with match. |
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30 Jun 2017, 12:26 |
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username 30 Jun 2017, 12:31
if you find a little spare time please post a working solution
thank you |
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30 Jun 2017, 12:31 |
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username 03 Jul 2017, 05:58
i modified a little a macro using the match directive, but i still am not able to get the results i need. i don't know how to use match to behave like an if statement
Code:
macro ext instr
{
macro instr op1,op2,op3
\{
instr op1,op2
if op3 eq
;nothing
else
match temp, op3
\\{
temp2 = temp mod 2
if temp2 = 0
match 0,temp2 ???? how to use match to make it execute on condition when temp2 = 0 ????
\\\{ exchange eaxul,ebxul \\\}
else
match 1,temp2 ???? how to use match to make it execute on condition when temp2 = 1 ????
\\\{ exchange ecxul,edxul \\\}
end if
\\}
end if
\}
} i tried match =0,temp2 ; match 0,=temp2 etc but i don't get the expected results if i don't get the use of match, please show me how to use it properly on my example so i need to evaluate and act accordingly (on conditions) some equations at the preprocessor layer, if i understand correctly. i saw the match directive example match =TRUE, DEBUG { include 'debug.inc' } but in my example it doesn't work if i use match =0,temp2 {....} |
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03 Jul 2017, 05:58 |
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revolution 04 Jul 2017, 07:21
You are still mixing if and match. Also all expressions are computed during the assembly stage so match has no knowledge of what value a numeric variable will have. Match only deals with string data, not numbers.
You can use equ and match together. Equ also deals with only string data. There is one preprocessor directive that can evaluate strings as numbers, that is rept. But it also generates string results of the numbers it evaluates. So you can use "rept x,temp mod 2" for example and x will then be equal to either "0" or "1" as a string. Then use match to see if x is 0 or 1. |
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04 Jul 2017, 07:21 |
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username 04 Jul 2017, 08:18
thank you for clarification
at the preprocessor layer is it possible to obtain a number without explicitly write it or get to it by calculations? something like $ ? |
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04 Jul 2017, 08:18 |
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revolution 04 Jul 2017, 08:26
You can't get the value of $ within the preprocessor. Only the assembly stage knows $.
If you use "temp = 4" then the preprocessor cannot know the value of temp. Even rept can't see the value of 4 because it is an expression, not an equate. You would have to use "temp equ 4" or the similar "define temp 4". Last edited by revolution on 04 Jul 2017, 11:01; edited 1 time in total |
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04 Jul 2017, 08:26 |
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username 04 Jul 2017, 08:41
nevermind, thanks again
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04 Jul 2017, 08:41 |
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