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Ali.Z 12 Oct 2019, 20:20
hey,
i hard-coded base32 encoding, currently it runs under ollydbg because i didnt write any gui code. (but i will once i write the decoder) however before writing the reversed code, i wanna improve this code and make it faster. many of you know i cant do math, and a lot of you helped me a lot already .. so thank you again in advance. (spent hours hard-coding the encoder  )
if you dont know the specification and standard of base32 encoding, then please visit this link. rfc 4648 Code: include 'win32a.inc' format pe gui 4.0 stack 1000h,1000h heap 1000h,1000h section '.text' code readable executable invoke GetModuleHandle,0 xor eax,eax or ecx,-1 mov edi,input repnz scasb not ecx dec ecx mov [input_length],ecx mov ebx,base_32_array mov esi,input mov edi,buffer cld _8bit: lodsb mov edx,eax and edx,7 ; extract first 3 bits mov [extracted_bits],edx ; store them shr eax,3 ; shift right by the number of extracted bits mov al,byte [ebx+eax] stosb _16bit: ; 3 bits left + 8 new bits = 11 bits - 5 bits(encode every 5bits) = 6 bits to shift dec ecx lodsb mov edx,eax and edx,3Fh ; extract first 6 bits xchg [extracted_bits],edx ; load previous bits, and store the new one shr eax,6 ; 2 bits left shl edx,2 ; shift by the number of bits left in eax or eax,edx ; combine bits 2,3,4 from edx with bits 0,1 in eax mov al,byte [ebx+eax] stosb or ecx,ecx jz _end .special: ; 6 bits left - 5 (encoding) = 1 bit to shift mov eax,[extracted_bits] mov edx,eax and edx,1 mov [extracted_bits],edx shr eax,1 mov al,byte [ebx+eax] stosb dec ecx or ecx,ecx jz _end _24bit: ; 1 bit left + 8 new bits = 9 bits - 5 = 4 bits to shift dec ecx lodsb mov edx,eax and edx,0Fh ; extract first 4 xchg [extracted_bits],edx ; edx = 1 bit shr eax,4 ; 4 bits left shl edx,4 ; shift by the number of bits left in eax or eax,edx ; combine bit 4 from edx with bits 0,1,2,3 in eax mov al,byte [ebx+eax] stosb or ecx,ecx jz _end _32bit: ; 4 bits left + 8 new bits = 12 - 5 = 7 bits to shift lodsb mov edx,eax and edx,7Fh ; extract first 7 xchg [extracted_bits],edx ; edx have 4 bits now shr eax,7 ; 1 bit left shl edx,1 ; shl by number of bits left in eax or eax,edx ; combine bits 1,2,3,4 with bit 0 in eax mov al,byte [ebx+eax] stosb .special: ; 7 bits left - 5 = 2 bits to shift mov eax,[extracted_bits] mov edx,eax and edx,3 mov [extracted_bits],edx shr eax,2 mov al,byte [ebx+eax] stosb dec ecx or ecx,ecx jz _end _40bit: ; 2 bits left + 8 new bits = 10 - 5 = 5 bits to shift lodsb mov edx,eax and edx,1Fh xchg [extracted_bits],edx shr eax,5 shl edx,3 or eax,edx mov al,byte [ebx+eax] stosb ..special: mov eax,[extracted_bits] mov al,byte [ebx+eax] stosb dec ecx or ecx,ecx jnz _8bit _end: mov ecx,[input_length] subtract: cmp ecx,5 jbe skip sub ecx,5 jmp subtract skip: cmp ecx,1 jz _6padding cmp ecx,2 jz _4padding cmp ecx,3 jz _3padding cmp ecx,4 jz _1padding int3 _6padding: mov ecx,6 mov eax,'=' rep stosb int3 _4padding: mov eax,[extracted_bits] shl eax,4 mov al,byte [ebx+eax] stosb mov ecx,4 mov eax,'=' rep stosb int3 _3padding: mov eax,[extracted_bits] shl eax,1 mov al,byte [ebx+eax] stosb mov ecx,3 mov eax,'=' rep stosb int3 _1padding: mov eax,[extracted_bits] shl eax,3 mov al,byte [ebx+eax] stosb mov byte [edi],'=' int3 section '.data' data readable writeable base_32_array db 'ABCDEFGHIJKLMNOPQRSTUVWXYZ234567',0 input db 'flat assembler',0 ; MZWGC5BAMFZXGZLNMJWGK4Q= buffer rb 1024 input_length dd 0 extracted_bits dd 0 section '.idata' import data readable library kernel32,'kernel32.dll' include 'api\kernel32.inc' section '.reloc' fixups data readable discardable _________________ Asm For Wise Humans |
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12 Oct 2019, 20:20 |
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Ali.Z 13 Oct 2019, 08:24
i wanted someone to help me to dynamically calculate how many bits i should extract and shift, but thats okay for now im still happy.
https://board.flatassembler.net/topic.php?p=209476#209476 revolution wrote: Division can be emulated by using repeated subtraction. Multiplication can be emulated by repeated addition. here im taking the total length of the input, and keep subtracting 5 to get the number of required padding. assuming the input was 1000 chars long, then i have to subtract 5 a lot .. im sure this can be done using div but im not sure what numbers i should put in there. Code: _end: mov ecx,[input_length] subtract: cmp ecx,5 jbe skip sub ecx,5 jmp subtract skip: cmp ecx,1 jz _6padding cmp ecx,2 jz _4padding cmp ecx,3 jz _3padding cmp ecx,4 jz _1padding int3 _________________ Asm For Wise Humans |
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13 Oct 2019, 08:24 |
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Ali.Z 13 Oct 2019, 14:12
yes, but ignore the encoder part for now.
i want to avoid subtracting by 5, and the solution is using DIV but i dont know what numbers i should put in would you mind fixing this thing for me. _________________ Asm For Wise Humans |
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13 Oct 2019, 14:12 |
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revolution 13 Oct 2019, 14:31
You can divide by five with this:
Code: mov eax,[length] xor edx,edx ;edx:eax is the numerator mov ecx,5 ;ecx is the denominator div ecx ;eax=floor(length/5) ;edx=remainder(length/5) |
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13 Oct 2019, 14:31 |
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Ali.Z 13 Oct 2019, 15:08
thanks rev, i didnt understand it .. but after running it under olly found that i should compare using edx after the div and ignore eax.
currently im hard-coding the decoder. _________________ Asm For Wise Humans |
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13 Oct 2019, 15:08 |
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Ali.Z 13 Oct 2019, 19:47
decoder is done, now the most boring thing is to design a gui window.
p.s. did some minor tweaks and got rid of extra code. _________________ Asm For Wise Humans |
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13 Oct 2019, 19:47 |
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