brainf

Başlatan messah, 10 Mart 2011 - 00:21:50

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messah

merhaba ;

araştırma yaparken brainfuck programlama diliyle karsılaştım ve ilgimi çekti. ilk bakıldığında çok basit bir dile benziyor fakat içine bakıldığında pek de öyle olmadığı gözüküyor. bu araştırmalar sırasında örnek kodlar aldım ".b" veya ".bf" uzantılı olabiliyorlar. ben bu kod parcalarını vim editörüyle kullanmak istiyorum acaba asıl derlerim ?

eribol

;; bf.asm: Copyright (C) 1999 Brian Raiter <breadbox@muppetlabs.com>
;; Licensed under the terms of the GNU General Public License, either
;; version 2 or (at your option) any later version.
;;
;; To build:
;; nasm -f bin -o bf bf.asm && chmod +x bf
;; To use:
;; bf < foo.b > foo && chmod +x foo

BITS 32

;; This is the size of the data area supplied to compiled programs.

%define arraysize 30000

;; For the compiler, the text segment is also the data segment. The
;; memory image of the compiler is inside the code buffer, and is
;; modified in place to become the memory image of the compiled
;; program. The area of memory that is the data segment for compiled
;; programs is not used by the compiler. The text and data segments of
;; compiled programs are really only different areas in a single
;; segment, from the system's point of view. Both the compiler and
;; compiled programs load the entire file contents into a single
;; memory segment which is both writeable and executable.

%define TEXTORG 0x45E9B000
%define DATAOFFSET 0x2000
%define DATAORG (TEXTORG + DATAOFFSET)

;; Here begins the file image.

org TEXTORG

;; At the beginning of the text segment is the ELF header and the
;; program header table, the latter consisting of a single entry. The
;; two structures overlap for a space of eight bytes. Nearly all
;; unused fields in the structures are used to hold bits of code.

;; The beginning of the ELF header.

db 0x7F, "ELF" ; ehdr.e_ident

;; The top(s) of the main compiling loop. The loop jumps back to
;; different positions, depending on how many bytes to copy into the
;; code buffer. After doing that, esi is initialized to point to the
;; epilog code chunk, a copy of edi (the pointer to the end of the
;; code buffer) is saved in ebp, the high bytes of eax are reset to
;; zero (via the exchange with ebx), and then the next character of
;; input is retrieved.

emitputchar: add esi, byte (putchar - decchar) - 4
emitgetchar: lodsd
emit6bytes: movsd
emit2bytes: movsb
emit1byte: movsb
compile: lea esi, [byte ecx + epilog - filesize]
xchg eax, ebx
cmp eax, 0x00030002 ; ehdr.e_type    (0x0002)
; ehdr.e_machine (0x0003)
mov ebp, edi ; ehdr.e_version
jmp short getchar

;; The entry point for the compiler (and compiled programs), and the
;; location of the program header table.

dd _start ; ehdr.e_entry
dd proghdr - $$ ; ehdr.e_phoff

;; The last routine of the compiler, called when there is no more
;; input. The epilog code chunk is copied into the code buffer. The
;; text origin is popped off the stack into ecx, and subtracted from
;; edi to determine the size of the compiled program. This value is
;; stored in the program header table, and then is moved into edx.
;; The program then jumps to the putchar routine, which sends the
;; compiled program to stdout before falling through to the epilog
;; routine and exiting.

eof: movsd ; ehdr.e_shoff
xchg eax, ecx
pop ecx
sub edi, ecx ; ehdr.e_flags
xchg eax, edi
stosd
xchg eax, edx
jmp short putchar ; ehdr.e_ehsize

;; 0x20 == the size of one program header table entry.

dw 0x20 ; ehdr.e_phentsize

;; The beginning of the program header table. 1 == PT_LOAD, indicating
;; that the segment is to be loaded into memory.

proghdr: dd 1 ; ehdr.e_phnum & phdr.p_type
; ehdr.e_shentsize
dd 0 ; ehdr.e_shnum & phdr.p_offset
; ehdr.e_shstrndx

;; (Note that the next four bytes, in addition to containing the first
;; two instructions of the bracket routine, also comprise the memory
;; address of the text origin.)

db 0 ; phdr.p_vaddr

;; The bracket routine emits code for the "[" instruction. This
;; instruction translates to a simple "jmp near", but the target of
;; the jump will not be known until the matching "]" is seen. The
;; routine thus outputs a random target, and pushes the location of
;; the target in the code buffer onto the stack.

bracket: mov al, 0xE9
inc ebp
push ebp ; phdr.p_paddr
stosd
jmp short emit1byte

;; This is where the size of the executable file is stored in the
;; program header table. The compiler updates this value just before
;; it outputs the compiled program. This is the only field in the two
;; headers that differs between the compiler and its compiled
;; programs. (While the compiler is reading input, the first byte of
;; this field is also used as an input buffer.)

filesize: dd compilersize ; phdr.p_filesz

;; The size of the program in memory. This entry creates an area of
;; bytes, arraysize in size, all initialized to zero, starting at
;; DATAORG.

dd DATAOFFSET + arraysize ; phdr.p_memsz

;; The code chunk for the "." instruction. eax is set to 4 to invoke
;; the write system call. ebx, the file handle to write to, is set to
;; 1 for stdout. ecx points to the buffer containing the bytes to
;; output, and edx equals the number of bytes to output. (Note that
;; the first byte of the first instruction, which is also the least
;; significant byte of the p_flags field, encodes to 0xB3. Having the
;; 2-bit set marks the memory containing the compiler, and its
;; compiled programs, as writeable.)

putchar: mov bl, 1 ; phdr.p_flags
mov al, 4
int 0x80 ; phdr.p_align

;; The epilog code chunk. After restoring the initialized registers,
;; eax and ebx are both zero. eax is incremented to 1, so as to invoke
;; the exit system call. ebx specifies the process's return value.

epilog: popa
inc eax
int 0x80

;; The code chunks for the ">", "<", "+", and "-" instructions.

incptr: inc ecx
decptr: dec ecx
incchar: inc byte [ecx]
decchar: dec byte [ecx]

;; The main loop of the compiler continues here, by obtaining the next
;; character of input. This is also the code chunk for the ","
;; instruction. eax is set to 3 to invoke the read system call. ebx,
;; the file handle to read from, is set to 0 for stdin. ecx points to
;; a buffer to receive the bytes that are read, and edx equals the
;; number of bytes to read.

getchar: mov al, 3
xor ebx, ebx
int 0x80

;; If eax is zero or negative, then there is no more input, and the
;; compiler proceeds to the eof routine.

or eax, eax
jle eof

;; Otherwise, esi is advanced four bytes (from the epilog code chunk
;; to the incptr code chunk), and the character read from the input is
;; stored in al, with the high bytes of eax reset to zero.

lodsd
mov eax, [ecx]

;; The compiler compares the input character with ">" and "<". esi is
;; advanced to the next code chunk with each failed test.

cmp al, '>'
jz emit1byte
inc esi
cmp al, '<'
jz emit1byte
inc esi

;; The next four tests check for the characters "+", ",", "-", and
;; ".", respectively. These four characters are contiguous in ASCII,
;; and so are tested for by doing successive decrements of eax.

sub al, '+'
jz emit2bytes
dec eax
jz emitgetchar
inc esi
inc esi
dec eax
jz emit2bytes
dec eax
jz emitputchar

;; The remaining instructions, "[" and "]", have special routines for
;; emitting the proper code. (Note that the jump back to the main loop
;; is at the edge of the short-jump range. Routines below here
;; therefore use this jump as a relay to return to the main loop;
;; however, in order to use it correctly, the routines must be sure
;; that the zero flag is cleared at the time.)

cmp al, '[' - '.'
jz bracket
cmp al, ']' - '.'
relay: jnz compile

;; The endbracket routine emits code for the "]" instruction, as well
;; as completing the code for the matching "[". The compiler first
;; emits "cmp dh, [ecx]" and the first two bytes of a "jnz near". The
;; location of the missing target in the code for the "[" instruction
;; is then retrieved from the stack, the correct target value is
;; computed and stored, and then the current instruction's jmp target
;; is computed and emitted.

endbracket: mov eax, 0x850F313A
stosd
lea esi, [byte edi - 8]
pop eax
sub esi, eax
mov [eax], esi
sub eax, edi
stosd
jmp short relay

;; This is the entry point, for both the compiler and its compiled
;; programs. The shared initialization code sets ecx to the beginning
;; of the array that is the compiled program's data area, and edx to
;; one. (This also clears the zero flag for the relay jump below.) The
;; registers are then saved on the stack, to be restored at the end.

_start:
xor ebx, ebx
mul ebx
mov ecx, DATAORG
inc edx
pusha

;; At this point, the compiler and its compiled programs diverge.
;; Although every compiled program includes all the code in this file
;; above this point, only the handful of instructions directly above
;; are actually used by both. This point is where the compiler begins
;; storing the generated code, so only the compiler sees the
;; instructions below. This routine first modifies ecx to contain
;; TEXTORG, which is stored on the stack, and then offsets it to point
;; to filesize. edi is set equal to codebuf, and then the compiler
;; enters the main loop.

codebuf:
mov ch, (TEXTORG >> 8) & 0xFF
push ecx
mov cl, filesize - $$
lea edi, [byte ecx + codebuf - filesize]
jmp short relay

;; Here ends the file image.

compilersize equ $ - $$
Bu kodları derlemelisiniz, asm derleyicisi ile. Ayrıca kendisi assembly tarzı bir dil. Tabii ne kadar gerçekçi bir dil bilmiyorum.

messah

yardımınız için teşekkürler. bayağı bir uğraşma gerektirecek gibi...

ekremsenturk

Synaptic'te BF ve BEEF var. BF bir yorumlayıcı, BEEF denemedim. Ve kendin kod yazacaksan, ASCII karakterlerini ezbere bilmen gerekiyor veya sürekli bir liste göz önünde bulundurmalısın.