Parsing 创建Brainfuck解析器，解析循环运算符的最佳方法是什么？
我正在创建一个Brainfuck解析器（用一种基本的方言）来最终创建一个解释器，但我意识到这并不像我最初想的那么简单。我的问题是，我需要一种在Brainfuck程序中准确解析匹配循环运算符的方法。这是一个示例程序：Parsing 创建Brainfuck解析器，解析循环运算符的最佳方法是什么？,parsing,loops,brainfuck,Parsing,Loops,Brainfuck,我正在创建一个Brainfuck解析器（用一种基本的方言）来最终创建一个解释器，但我意识到这并不像我最初想的那么简单。我的问题是，我需要一种在Brainfuck程序中准确解析匹配循环运算符的方法。这是一个示例程序： ,>,>++++++++[<------<------>>-] <<[>[>+>+<<-]>>[<<+>>-]<<<-] >>>+++
,>,>++++++++[<------<------>>-]
<<[>[>+>+<<-]>>[<<+>>-]<<<-]
>>>++++++[<++++++++>-],<.>.
，>，>++++++++[-]
+>+[-]++++++[-],.
编辑：非常欣赏任何语言的示例代码。您是否考虑过使用堆栈数据结构来记录“跳转点”（即指令指针的位置） 因此，基本上，每次遇到“[”都会推送指令指针在此堆栈上的当前位置。每当遇到“]”时，都会将指令指针重置为当前位于堆栈顶部的值。循环完成后，将其从堆栈中弹出
这里是一个C++的例子，里面有100个存储单元。代码以递归方式处理嵌套循环，虽然没有细化，但应该说明这些概念
char cells[100] = {0}; // define 100 memory cells
char* cell = cells; // set memory pointer to first cell
char* ip = 0; // define variable used as "instruction pointer"
void interpret(static char* program, int* stack, int sp)
{
int tmp;
if(ip == 0) // if the instruction pointer hasn't been initialized
ip = program; // now would be a good time
while(*ip) // this runs for as long as there is valid brainF**k 'code'
{
if(*ip == ',')
*cell = getch();
else if(*ip == '.')
putch(*cell);
else if(*ip == '>')
cell++;
else if(*ip == '<')
cell--;
else if(*ip == '+')
*cell = *cell + 1;
else if(*ip == '-')
*cell = *cell - 1;
else if(*ip == '[')
{
stack[sp+1] = ip - program;
*ip++;
while(*cell != 0)
{
interpret(program, stack, sp + 1);
}
tmp = sp + 1;
while((tmp >= (sp + 1)) || *ip != ']')
{
*ip++;
if(*ip == '[')
stack[++tmp] = ip - program;
else if(*ip == ']')
tmp--;
}
}
else if(*ip == ']')
{
ip = program + stack[sp] + 1;
break;
}
*ip++; // advance instruction
}
}
int _tmain(int argc, _TCHAR* argv[])
{
int stack[100] = {0}; // use a stack of 100 levels, modeled using a simple array
interpret(",>,>++++++++[<------<------>>-]<<[>[>+>+<<-]>>[<<+>>-]<<<-]>>>++++++[<++++++++>-],<.>.", stack, 0);
return 0;
}
char cells[100] = {0};
void interpret(static char* program)
{
int cnt; // cnt is a counter that is going to be used
// only when parsing 0-loops
int stack[100] = {0}; // create a stack, 100 levels deep - modeled
// using a simple array - and initialized to 0
int sp = 0; // sp is going to be used as a 'stack pointer'
char* ip = program; // ip is going to be used as instruction pointer
// and it is initialized at the beginning or program
char* cell = cells; // cell is the pointer to the 'current' memory cell
// and as such, it is initialized to the first
// memory cell
while(*ip) // as long as ip point to 'valid code' keep going
{
if(*ip == ',')
*cell = getch();
else if(*ip == '.')
putch(*cell);
else if(*ip == '>')
cell++;
else if(*ip == '<')
cell--;
else if(*ip == '+')
*cell = *cell + 1;
else if(*ip == '-')
*cell = *cell - 1;
else if(*ip == '[')
{
if(stack[sp] != ip - program)
stack[++sp] = ip - program;
*ip++;
if(*cell != 0)
continue;
else
{
cnt = 1;
while((cnt > 0) || *ip != ']')
{
*ip++;
if(*ip == '[')
cnt++;
else if(*ip == ']')
cnt--;
}
sp--;
}
}else if(*ip == ']')
{
ip = program + stack[sp];
continue;
}
*ip++;
}
}
int _tmain(int argc, _TCHAR* argv[])
{
// define our program code here..
char *prg = ",>++++++[<-------->-],[<+>-]<.";
interpret(prg);
return 0;
}
char单元格[100]={0}；
无效解释（静态字符*程序）
{
int cnt；//cnt是将要使用的计数器
//仅在分析0-1循环时
int stack[100]={0}；//创建一个堆栈，100层深-建模
//使用简单数组-并初始化为0
int sp=0；//sp将用作“堆栈指针”
char*ip=program；//ip将用作指令指针
//它在程序的开始或结束时初始化
char*cell=cells；//cell是指向“当前”内存单元的指针
//因此，它被初始化为第一个
//存储单元
while（*ip）//只要ip指向“有效代码”，则继续运行
{
如果（*ip=='，'）
*cell=getch（）；
否则，如果（*ip=='。）
putch（*单元）；
否则，如果（*ip=='>'））
cell++；
否则，如果（*ip=='每次找到“[”，请在堆栈上推送当前位置（或另一个“标记”标记或“上下文”）。当您越过“]”时，您处于循环的末尾，可以从堆栈中弹出标记标记
因为在BF中，“[”已经检查了一个条件，可能需要跳过“]”，所以您可能需要一个标志，指示在当前循环上下文中跳过指令。我没有示例代码，但是
我可能会尝试使用堆栈，并使用如下算法：
（执行指令流）
遭遇[
如果指针==0，则继续读取，直到遇到“]”，并且在找到它之前不要执行任何指令。转到步骤1
如果指针！=0，则将该位置推到堆栈上
继续执行指令
如果您遇到一个问题]
如果指针==0，则从堆栈中弹出[并继续（转到步骤1）
如果指针！=0，查看堆栈顶部，然后转到该位置。（转到步骤5）
其他海报描述的Python 3.0堆栈算法示例：
program = """
,>,>++++++++[<------<------>>-]
<<[>[>+>+<<-]>>[<<+>>-]<<<-]
>>>++++++[<++++++++>-],<.>.
"""
def matching_brackets(program):
stack = []
for p, c in enumerate(program, start=1):
if c == '[':
stack.append(p)
elif c == ']':
yield (stack.pop(), p)
print(list(matching_brackets(''.join(program.split()))))
program=”“”
,>,>++++++++[-]
+>+[-]++++++[-],.
"""
def匹配_括号（程序）：
堆栈=[]
对于枚举中的p，c（程序，开始=1）：
如果c=='['：
stack.append（p）
elif c==']'：
产量（stack.pop（），p）
打印（列表（匹配的括号（“”.join（program.split（）））
（老实说，这只找到匹配的括号。我不知道brainf*ck，所以我不知道下一步该怎么做。）很有趣，就在几天前，我正在用Java编写一个brainf*ck解释器
我遇到的一个问题是，当时对命令的解释不够充分，没有提到关于嵌套循环的部分
基本上，为了总结这个问题，官方页面说，当指令是[
且当前内存位置是0
，然后跳转到下一个]
。正确的行为是跳转到相应的]
，而不是下一个
实现此行为的一种方法是跟踪嵌套级别。我通过使用一个跟踪嵌套级别的计数器来实现这一点
以下是解释器主循环的一部分：
do {
if (inst[pc] == '>') { ... }
else if (inst[pc] == '<') { ... }
else if (inst[pc] == '+') { ... }
else if (inst[pc] == '-') { ... }
else if (inst[pc] == '.') { ... }
else if (inst[pc] == ',') { ... }
else if (inst[pc] == '[') {
if (memory[p] == 0) {
int nesting = 0;
while (true) {
++pc;
if (inst[pc] == '[') {
++nesting;
continue;
} else if (nesting > 0 && inst[pc] == ']') {
--nesting;
continue;
} else if (inst[pc] == ']' && nesting == 0) {
break;
}
}
}
}
else if (inst[pc] == ']') {
if (memory[p] != 0) {
int nesting = 0;
while (true) {
--pc;
if (inst[pc] == ']') {
++nesting;
continue;
} else if (nesting > 0 && inst[pc] == '[') {
--nesting;
continue;
} else if (inst[pc] == '[' && nesting == 0) {
break;
}
}
}
}
} while (++pc < inst.length);
do{
如果（inst[pc]=='>'）{…}
否则，（VB.NET）[PC]=＝`p>，这里是我在C++中作为例子给出的相同的代码，但是移植到了加里。我决定把它贴在这里，因为加里提到他试图用基本的方言编写他的解析器。
Public cells(100) As Byte
Sub interpret(ByVal prog As String)
Dim program() As Char
program = prog.ToCharArray() ' convert the input program into a Char array
Dim cnt As Integer = 0 ' a counter to be used when skipping over 0-loops
Dim stack(100) As Integer ' a simple array to be used as stack
Dim sp As Integer = 0 ' stack pointer (current stack level)
Dim ip As Integer = 0 ' Instruction pointer (index of current instruction)
Dim cell As Integer = 0 ' index of current memory
While (ip < program.Length) ' loop over the program
If (program(ip) = ",") Then
cells(cell) = CByte(AscW(Console.ReadKey().KeyChar))
ElseIf (program(ip) = ".") Then
Console.Write("{0}", Chr(cells(cell)))
ElseIf (program(ip) = ">") Then
cell = cell + 1
ElseIf (program(ip) = "<") Then
cell = cell - 1
ElseIf (program(ip) = "+") Then
cells(cell) = cells(cell) + 1
ElseIf (program(ip) = "-") Then
cells(cell) = cells(cell) - 1
ElseIf (program(ip) = "[") Then
If (stack(sp) <> ip) Then
sp = sp + 1
stack(sp) = ip
End If
ip = ip + 1
If (cells(cell) <> 0) Then
Continue While
Else
cnt = 1
While ((cnt > 0) Or (program(ip) <> "]"))
ip = ip + 1
If (program(ip) = "[") Then
cnt = cnt + 1
ElseIf (program(ip) = "]") Then
cnt = cnt - 1
End If
End While
sp = sp - 1
End If
ElseIf (program(ip) = "]") Then
ip = stack(sp)
Continue While
End If
ip = ip + 1
End While
End Sub
Sub Main()
' invoke the interpreter
interpret(",>++++++[<-------->-],[<+>-]<.")
End Sub
公共单元（100）作为字节
子解释（ByVal prog作为字符串）
Dim program（）作为字符
program=prog.tocharray（）'将输入程序转换为字符数组
Dim cnt As Integer=0'跳过0个循环时使用的计数器
Dim stack（100）As Integer'用作堆栈的简单数组
Dim sp As Integer=0'堆栈指针（当前堆栈级别）
Dim ip作为整数=0'指令指针（当前指令的索引）
将单元格设置为整数=当前内存的0'索引
而（ipPublic cells(100) As Byte
Sub interpret(ByVal prog As String)
Dim program() As Char
program = prog.ToCharArray() ' convert the input program into a Char array
Dim cnt As Integer = 0 ' a counter to be used when skipping over 0-loops
Dim stack(100) As Integer ' a simple array to be used as stack
Dim sp As Integer = 0 ' stack pointer (current stack level)
Dim ip As Integer = 0 ' Instruction pointer (index of current instruction)
Dim cell As Integer = 0 ' index of current memory
While (ip < program.Length) ' loop over the program
If (program(ip) = ",") Then
cells(cell) = CByte(AscW(Console.ReadKey().KeyChar))
ElseIf (program(ip) = ".") Then
Console.Write("{0}", Chr(cells(cell)))
ElseIf (program(ip) = ">") Then
cell = cell + 1
ElseIf (program(ip) = "<") Then
cell = cell - 1
ElseIf (program(ip) = "+") Then
cells(cell) = cells(cell) + 1
ElseIf (program(ip) = "-") Then
cells(cell) = cells(cell) - 1
ElseIf (program(ip) = "[") Then
If (stack(sp) <> ip) Then
sp = sp + 1
stack(sp) = ip
End If
ip = ip + 1
If (cells(cell) <> 0) Then
Continue While
Else
cnt = 1
While ((cnt > 0) Or (program(ip) <> "]"))
ip = ip + 1
If (program(ip) = "[") Then
cnt = cnt + 1
ElseIf (program(ip) = "]") Then
cnt = cnt - 1
End If
End While
sp = sp - 1
End If
ElseIf (program(ip) = "]") Then
ip = stack(sp)
Continue While
End If
ip = ip + 1
End While
End Sub
Sub Main()
' invoke the interpreter
interpret(",>++++++[<-------->-],[<+>-]<.")
End Sub
#include <stdio.h>
char *S[9999], P[9999], T[9999],
**s=S, *p=P, *t=T, c, x;
int main() {
fread(p, 1, 9999, stdin);
for (; c=*p; ++p) {
if (c == ']') {
if (!x)
if (*t) p = *(s-1);
else --s;
else --x;
} else if (!x) {
if (c == '[')
if (*t) *(s++) = p;
else ++x;
}
if (c == '<') t--;
if (c == '>') t++;
if (c == '+') ++*t;
if (c == '-') --*t;
if (c == ',') *t = getchar();
if (c == '.') putchar(*t);
}
}
}
#include <assert.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <sys/mman.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
void error(char *msg) {
fprintf(stderr, "Error: %s\n", msg);
}
enum { MEMSIZE = 30000 };
char *mem;
char *ptr;
char *prog;
size_t progsize;
int init(char *progname) {
int f,r;
struct stat fs;
ptr = mem = calloc(MEMSIZE, 1);
f = open(progname, O_RDONLY);
assert(f != -1);
r = fstat(f, &fs);
assert(r == 0);
prog = mmap(NULL, progsize = fs.st_size, PROT_READ, MAP_PRIVATE, f, 0);
assert(prog != NULL);
return 0;
}
int findmatch(int ip, char src){
char *p="[]";
int dir[]= { 1, -1 };
int i;
int defer;
i = strchr(p,src)-p;
ip+=dir[i];
for (defer=dir[i]; defer!=0; ip+=dir[i]) {
if (ip<0||ip>=progsize) error("mismatch");
char *q = strchr(p,prog[ip]);
if (q) {
int j = q-p;
defer+=dir[j];
}
}
return ip;
}
int run() {
int ip;
for(ip = 0; ip>=0 && ip<progsize; ip++)
switch(prog[ip]){
case '>': ++ptr; break;
case '<': --ptr; break;
case '+': ++*ptr; break;
case '-': --*ptr; break;
case '.': putchar(*ptr); break;
case ',': *ptr=getchar(); break;
case '[': /*while(*ptr){*/
if (!*ptr) ip=findmatch(ip,'[');
break;
case ']': /*}*/
if (*ptr) ip=findmatch(ip,']');
break;
}
return 0;
}
int cleanup() {
free(mem);
ptr = NULL;
return 0;
}
int main(int argc, char *argv[]) {
init(argc > 1? argv[1]: NULL);
run();
cleanup();
return 0;
}
package interpreter;
import java.awt.event.ActionListener;
import javax.swing.JTextPane;
public class Brainfuck {
final int tapeSize = 0xFFFF;
int tapePointer = 0;
int[] tape = new int[tapeSize];
int inputCounter = 0;
ActionListener onUpdateTape;
public Brainfuck(byte[] input, String code, boolean debugger,
JTextPane output, ActionListener onUpdate) {
onUpdateTape = onUpdate;
if (debugger) {
debuggerBF(input, code, output);
} else {
cleanBF(input, code, output);
}
}
private void debuggerBF(byte[] input, String code, JTextPane output) {
for (int i = 0; i < code.length(); i++) {
onUpdateTape.actionPerformed(null);
switch (code.charAt(i)) {
case '+': {
tape[tapePointer]++;
break;
}
case '-': {
tape[tapePointer]--;
break;
}
case '<': {
tapePointer--;
break;
}
case '>': {
tapePointer++;
break;
}
case '[': {
if (tape[tapePointer] == 0) {
int nesting = 0;
while (true) {
++i;
if (code.charAt(i) == '[') {
++nesting;
continue;
} else if (nesting > 0 && code.charAt(i) == ']') {
--nesting;
continue;
} else if (code.charAt(i) == ']' && nesting == 0) {
break;
}
}
}
break;
}
case ']': {
if (tape[tapePointer] != 0) {
int nesting = 0;
while (true) {
--i;
if (code.charAt(i) == ']') {
++nesting;
continue;
} else if (nesting > 0 && code.charAt(i) == '[') {
--nesting;
continue;
} else if (code.charAt(i) == '[' && nesting == 0) {
break;
}
}
}
break;
}
case '.': {
output.setText(output.getText() + (char) (tape[tapePointer]));
break;
}
case ',': {
tape[tapePointer] = input[inputCounter];
inputCounter++;
break;
}
}
}
}
private void cleanBF(byte[] input, String code, JTextPane output) {
for (int i = 0; i < code.length(); i++) {
onUpdateTape.actionPerformed(null);
switch (code.charAt(i)) {
case '+':{
tape[tapePointer]++;
break;
}
case '-':{
tape[tapePointer]--;
break;
}
case '<':{
tapePointer--;
break;
}
case '>':{
tapePointer++;
break;
}
case '[': {
if (tape[tapePointer] == 0) {
int nesting = 0;
while (true) {
++i;
if (code.charAt(i) == '[') {
++nesting;
continue;
} else if (nesting > 0 && code.charAt(i) == ']') {
--nesting;
continue;
} else if (code.charAt(i) == ']' && nesting == 0) {
break;
}
}
}
break;
}
case ']': {
if (tape[tapePointer] != 0) {
int nesting = 0;
while (true) {
--i;
if (code.charAt(i) == ']') {
++nesting;
continue;
} else if (nesting > 0 && code.charAt(i) == '[') {
--nesting;
continue;
} else if (code.charAt(i) == '[' && nesting == 0) {
break;
}
}
}
break;
}
case '.':{
output.setText(output.getText()+(char)(tape[tapePointer]));
break;
}
case ',':{
tape[tapePointer] = input[inputCounter];
inputCounter++;
break;
}
}
}
}
public int[] getTape() {
return tape;
}
public void setTape(int[] tape) {
this.tape = tape;
}
public void editTapeValue(int counter, int value) {
this.tape[counter] = value;
}
}
case '[': {
if (tape[tapePointer] == 0) {
int nesting = 0;
while (true) {
++i;
if (code.charAt(i) == '[') {
++nesting;
continue;
}
else if (nesting > 0 && code.charAt(i) == ']') {
--nesting;
continue;
}
else if (code.charAt(i) == ']' && nesting == 0) {
break;
}
}
}
break;
}
case ']': {
if (tape[tapePointer] != 0) {
int nesting = 0;
while (true) {
--i;
if (code.charAt(i) == ']') {
++nesting;
continue;
}
else if (nesting > 0 && code.charAt(i) == '[') {
--nesting;
continue;
}
else if (code.charAt(i) == '[' && nesting == 0) {
break;
}
}
}
break;
}