Zestaw x86 (w systemie Win32)
„SPEED!” Wydaje się tutaj niezwykle ważne i wszyscy wiemy, że pod tym względem nic nie przebije języka asemblera. Zróbmy to w asemblerze!
Jest to implementacja języka w asemblerze x86 (w składni NASM), z liczbami przechowywanymi i interpretowanymi jako 32-bitowe liczby całkowite bez znaku, bezpośrednio przy użyciu natywnego stosu x86. Niedopełnienie stosu i przepełnienie podczas dowolnej operacji arytmetycznej (lub dzielenia przez zero) jest błędem środowiska wykonawczego, kończącym program z komunikatem o błędzie.
global _start
extern _GetCommandLineA@0
extern _GetStdHandle@4
extern _CreateFileA@28
extern _GetFileSize@8
extern _LocalAlloc@8
extern _ReadFile@20
extern _CloseHandle@4
extern _WriteFile@20
section .text
; ---------------------------------------------------------------------------------------
; Initialization
; ---------------------------------------------------------------------------------------
_start:
; Retrieve command line
CALL _GetCommandLineA@0
; Skip argv[0]
MOV ESI, EAX
XOR EAX, EAX
skipuntilspace:
MOV AL, [ESI]
INC ESI
TEST EAX, EAX
JE missingparam
CMP EAX, ' '
JNE skipuntilspace
INC ESI
; Open the file
PUSH 0
PUSH 80h
PUSH 3
PUSH 0
PUSH 1
PUSH 80000000h
PUSH ESI
CALL _CreateFileA@28
CMP EAX, -1
JE cannotopenfile
; Get its size
PUSH EAX
PUSH 0
PUSH EAX
CALL _GetFileSize@8
PUSH EAX
; Allocate memory buffer
PUSH EAX
PUSH 0
CALL _LocalAlloc@8
TEST EAX, EAX
MOV ESI, EAX
JZ outofmemory
POP ECX
POP EAX
PUSH EAX
; Store end-of-program pointer
MOV [programend], ESI
ADD [programend], ECX
; Read the file contents
PUSH 0
PUSH buff
PUSH ECX
PUSH ESI
PUSH EAX
CALL _ReadFile@20
TEST EAX, EAX
JZ cannotopenfile
; Close the file
CALL _CloseHandle@4
; ---------------------------------------------------------------------------------------
; Main loop of the interpreter
; ---------------------------------------------------------------------------------------
; Store the end of stack into EBP
MOV EBP, ESP
; Push an initial 0 onto the stack
XOR EAX, EAX
PUSH EAX
mainloop:
; Load the next opcode, if not end of program
XOR EAX, EAX
CMP ESI, [programend]
MOV AL, [ESI]
JAE endloop
LEA ESI, [ESI+1]
; Check if the opcode is valid
CMP EAX, (maxop - opcodetable) / 8
JA fault_invalidopcode
; Check for required stack space
MOV ECX, [opcodetable + 8 * EAX + 4]
LEA EDI, [ESP + ECX]
CMP EDI, EBP
JA fault_stackunderflow
; Jump to the respective opcode handler
MOV EAX, [opcodetable + 8 * EAX]
JMP EAX
; ---------------------------------------------------------------------------------------
; Implementation of the specific operations
; ---------------------------------------------------------------------------------------
; ************** CAT 0000 (0): Concatenate (Combine top two numbers in a stack as if they were a string. ex: 12,5 -> 125)
op_concatenate:
POP EBX
POP EAX
MOV ECX, EAX
MOV EDI, 10
concat_loop:
XOR EDX, EDX
SHL EBX, 1
DIV EDI
LEA EBX, [4 * EBX + EBX]
TEST EAX, EAX
JNZ concat_loop
ADD EBX, ECX
PUSH EBX
JMP mainloop
; ************** INC 0001 (1): Increment (Add 1 to the number on the top of the stack)
op_increment:
POP EAX
ADD EAX, 1
PUSH EAX
JNC mainloop
JMP fault_intoverflow
; ************** DEC 0010 (2): Decrement (Subtract one from the number at the top of the stack)
op_decrement:
POP EAX
SUB EAX, 1
PUSH EAX
JNC mainloop
JMP fault_intoverflow
; ************** MUL 0011 (3): Multiply (Multiply the top two numbers in the stack)
op_multiply:
POP EAX
POP EDX
MUL EDX
TEST EDX, EDX
PUSH EAX
JZ mainloop
JMP fault_intoverflow
; ************** DIV 0100 (4): Divide (Divide the 2nd-to-top number by the top number on the stack)
op_divide:
POP ECX
TEST ECX, ECX
POP EAX
JZ fault_dividebyzero
XOR EDX, EDX
DIV ECX
PUSH EAX
JMP mainloop
; ************** MOD 0101 (5): Add (Add the top two numbers on the stack)
op_add:
POP EAX
ADD [ESP], EAX
JNC mainloop
JMP fault_intoverflow
; ************** SUB 0110 (6): Subtract (Subtract the top number on the stack from the one below it)
op_subtract:
POP EAX
SUB [ESP], EAX
JNC mainloop
JMP fault_intoverflow
; ************** EXP 0111 (7): Exponent (Calculate the second-to-top number to the power of the top number)
op_exponent:
POP ECX
POP EBX
MOV EAX, 1
exploop:
TEST ECX, 1
JZ expnomult
MUL EBX
TEST EDX, EDX
JNZ fault_intoverflow
expnomult:
SHR ECX, 1
JZ expdone
XCHG EAX, EBX
MUL EAX
TEST EDX, EDX
XCHG EAX, EBX
JZ exploop
JMP fault_intoverflow
expdone:
PUSH EAX
JMP mainloop
; ************** MOD 1000 (8): Modulus: (Find the second-to-top number modulo the top one)
op_modulus:
POP ECX
TEST ECX, ECX
POP EAX
JZ fault_dividebyzero
XOR EDX, EDX
IDIV ECX
PUSH EDX
JMP mainloop
; ************** ROR 1001 (9): Rotate Right (Shift the stack down one. The number on the bottom is now on the top)
op_rotright:
MOV EAX, [EBP - 4]
LEA ECX, [EBP - 4]
SUB ECX, ESP
MOV EDX, ESI
SHR ECX, 2
LEA EDI, [EBP - 4]
LEA ESI, [EBP - 8]
STD
REP MOVSD
MOV [ESP], EAX
CLD
MOV ESI, EDX
JMP mainloop
; ************** ROL 1010 (A): Rotate Left (Shift the stack up one. The number on the top is now on the bottom)
op_rotleft:
MOV EAX, [ESP]
LEA ECX, [EBP - 4]
SUB ECX, ESP
MOV EDX, ESI
SHR ECX, 2
LEA ESI, [ESP + 4]
MOV EDI, ESP
REP MOVSD
MOV [EBP - 4], EAX
MOV ESI, EDX
JMP mainloop
; ************** DUP 1011 (B): Duplicate (Copy the top number so that it appears twice. ex: 4,1 becomes 4,1,1)
op_duplicate:
PUSH DWORD [ESP]
JMP mainloop
; ************** DU2 1100 (C): Double Duplicate (Copy the top two numbers on the stack. ex: 4,1,2 becomes 4,1,2,1,2)
op_dblduplicate:
PUSH DWORD [ESP+4]
PUSH DWORD [ESP+4]
JMP mainloop
; ************** SWP 1101 (D): Swap (Swap the top two numbers on the stack. ex: 4,1,2 becomes 4,2,1)
op_swap:
POP EAX
POP EDX
PUSH EAX
PUSH EDX
JMP mainloop
; ************** SW2 1110 (E): Double Swap (Swap the top two numbers with two below them.ex: 1,2,3,4,5 becomes 1,4,5,2,3)
op_dblswap:
POP EAX
POP EBX
POP ECX
POP EDX
PUSH EBX
PUSH EAX
PUSH EDX
PUSH ECX
JMP mainloop
; ************** POP 1111 (F): Delete/Pop (Remove the number at the top of the stack)
op_pop:
POP EAX
JMP mainloop
; ---------------------------------------------------------------------------------------
; End of the program: print out the resulting stack and exit
; ---------------------------------------------------------------------------------------
endloop:
MOV ESI, ESP
printloop:
CMP ESI, EBP
JNB exit
MOV EAX, [ESI]
MOV EBX, ESI
PUSH EBX
CALL printnum
POP EBX
LEA ESI, [EBX + 4]
JMP printloop
exit:
MOV ESP, EBP
;POP EAX
XOR EAX, EAX
RET
; ---------------------------------------------------------------------------------------
; Faults
; ---------------------------------------------------------------------------------------
fault_invalidopcode:
MOV EAX, err_invalidopcode
JMP fault
fault_stackunderflow:
MOV EAX, err_stackunderflow
JMP fault
fault_dividebyzero:
MOV EAX, err_dividebyzero
JMP fault
fault_intoverflow:
MOV EAX, err_intoverflow
JMP fault
fault:
CALL print
MOV EAX, crlf
CALL print
MOV ESP, EBP
MOV EAX, 1
RET
missingparam:
MOV EAX, err_missingparameter
JMP fault
cannotopenfile:
MOV EAX, err_cannotopenfile
JMP fault
outofmemory:
MOV EAX, err_outofmemory
JMP fault
; ---------------------------------------------------------------------------------------
; Helper functions
; ---------------------------------------------------------------------------------------
printnum:
MOV EBX, 10
CALL printnumrec
MOV EAX, crlf
JMP print
printnumrec:
PUSH EAX
PUSH EDX
XOR EDX, EDX
DIV EBX
TEST EAX, EAX
JZ printnumend
CALL printnumrec
printnumend:
MOV EAX, EDX
CALL printdigit
POP EDX
POP EAX
RET
printdigit:
ADD EAX, '0'
MOV [printbuff], EAX
MOV EAX, printbuff
JMP print
print:
MOV ESI, EAX
PUSH 0
PUSH buff
CALL strlen
PUSH EAX
PUSH ESI
PUSH -11
CALL _GetStdHandle@4
PUSH EAX
CALL _WriteFile@20
RET
strlen:
XOR ECX, ECX
strlen_loop:
CMP BYTE [ESI+ECX], 0
JE strlen_end
LEA ECX, [ECX+1]
JMP strlen_loop
strlen_end:
MOV EAX, ECX
RET
; ---------------------------------------------------------------------------------------
; Data
; ---------------------------------------------------------------------------------------
section .data
; Table of opcode handlers and required stack space (in bytes, i.e. 4*operands)
opcodetable:
DD op_concatenate, 8
DD op_increment, 4
DD op_decrement, 4
DD op_multiply, 8
DD op_divide, 8
DD op_add, 8
DD op_subtract, 8
DD op_exponent, 8
DD op_modulus, 8
DD op_rotright, 0
DD op_rotleft, 0
DD op_duplicate, 4
DD op_dblduplicate, 8
DD op_swap, 8
DD op_dblswap, 16
DD op_pop, 4
maxop:
crlf DB 13, 10, 0
err_invalidopcode DB "Invalid opcode", 0
err_stackunderflow DB "Stack underflow", 0
err_dividebyzero DB "Division by zero", 0
err_intoverflow DB "Integer overflow", 0
err_missingparameter: DB "Missing parameter: Use nexlang file.bin", 0
err_cannotopenfile: DB "Unable to open input file", 0
err_outofmemory: DB "Not enough memory", 0
section .bss
programend RESD 1
printbuff RESD 1
buff RESD 1
Aby to skompilować, użyj czegoś takiego
nasm.exe -fwin32 nexlang.asm
ld -o nexlang.exe -e _start nexlang.obj -s -lkernel32
Program otrzymuje nazwę pliku binarnego zawierającego program w wierszu poleceń (np nexlang.exe testprg.bin.). Po zakończeniu drukuje końcową zawartość stosu na standardowe wyjście w formacie czytelnym dla człowieka.
Aby pomóc w testowaniu, zapisz następujące elementy w nex.def:
%define CAT DB 00h
%define INC DB 01h
%define DEC DB 02h
%define MUL DB 03h
%define DIV DB 04h
%define ADD DB 05h
%define SUB DB 06h
%define EXP DB 07h
%define MOD DB 08h
%define ROR DB 09h
%define ROL DB 0Ah
%define DUP DB 0Bh
%define DU2 DB 0Ch
%define SWP DB 0Dh
%define SW2 DB 0Eh
%define POP DB 0Fh
A następnie napisz swoje programy NEX („nieistniejące”, jak podano w tytule pytania), używając wyżej zdefiniowanej mnemoniki i skompiluj z czymś w rodzaju
nasm.exe -p nex.def -o prg.bin prg.nex
Np. W przypadku oryginalnego przypadku testowego użyj prg.nex:
INC ; 1
INC ; 2
INC ; 3
INC ; 4
DUP ; 4 4
DU2 ; 4 4 4 4
ADD ; 8 4 4
DU2 ; 8 4 8 4 4
ADD ; 12 8 4 4
DUP ; 12 12 8 4 4
ROR ; 4 12 12 8 4
ADD ; 16 12 8 4
I na koniec, w przypadku wyzwania „2014” użyj następującego 14-bajtowego programu NEX:
DUP ; 0 0
DUP ; 0 0 0
INC ; 1 0 0
INC ; 2 0 0
SWP ; 0 2 0
CAT ; 20 0
SWP ; 0 20
INC ; 1 20
DUP ; 1 1 20
INC ; 2 1 20
INC ; 3 1 20
INC ; 4 1 20
CAT ; 14 20
CAT ; 2014