Creating a Brainfuck parser, whats the best method of parsing loop operators?

半世苍凉 提交于 2019-12-05 09:59:11

Have you considered using a Stack data structure to record "jump points" (i.e. the location of the instruction pointer).

So basically, every time you encounter a "[" you push the current location of the instruction pointer on this stack. Whenever you encounter a "]" you reset the instruction pointer to the value that's currently on the top of the stack. When a loop is complete, you pop it off the stack.

Here is an example in C++ with 100 memory cells. The code handles nested loops recursively and although it is not refined it should illustrate the concepts..

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;
}

EDIT I just went over the code again and I realized there was a bug in the while loop that would 'skip' parsed loops if the value of the pointer is 0. This is where I made the change:

while((tmp >= (sp + 1)) || *ip != ']')     // the bug was tmp > (sp + 1)
{
ip++;
if(*ip == '[')
    stack[++tmp] = ip - program;
else if(*ip == ']')
    tmp--;
}

Below is an implementation of the same parser but without using recursion:

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;
}

The canonical method for parsing a context-free grammar is to use a stack. Anything else and you're working too hard and risking correctness.

You may want to use a parser generator like cup or yacc, as a lot of the dirty work is done for you, but with a language as simple as BF, it may be overkill.

Interesting enough, just a couple days ago, I was writing a brainf*ck interpreter in Java.

One of the issues I was having was that the explanation of the commands at the official page was insufficient, and did not mention the part about nested loops. The Wikipedia page on Brainf*ck has a Commands subsection which describes the correct behavior.

Basically to summarize the problem, the official page says when an instruction is a [ and the current memory location is 0, then jump to the next ]. The correct behavior is to jump to the corresponding ], not the next one.

One way to achieve this behavior is to keep track of the level of nesting. I ended up implementing this by having a counter which kept track of the nesting level.

The following is part of the interpreter's main loop:

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);

Here is the legend for the variable names:

  • memory -- the memory cells for the data.
  • p -- pointer to the current memory cell location.
  • inst -- an array holding the instructions.
  • pc -- program counter; points to the current instruction.
  • nesting -- level of the nesting of the current loop. nesting of 0 means that the current location is not in a nested loop.

Basically, when a loop opening [ is encountered, the current memory location is checked to see if the value is 0. If that is the case, a while loop is entered to jump to the corresponding ].

The way the nesting is handled is as follows:

  1. If an [ is encountered while seeking for the corresponding loop closing ], then the nesting variable is incremented by 1 in order to indicate that we have entered a nested loop.

  2. If an ] is encountered, and:

    a. If the nesting variable is greater than 0, then the nesting variable is decremented by 1 to indicate that we've left a nested loop.

    b. If the nesting variable is 0, then we know that the end of the loop has been encountered, so seeking the end of the loop in the while loop is terminated by executing a break statement.

Now, the next part is to handle the closing of the loop by ]. Similar to the opening of the loop, it will use the nesting counter in order to determine the current nesting level of the loop, and try to find the corresponding loop opening [.

This method may not be the most elegant way to do things, but it seems like it is resource-friendly because it only requires one extra variable to use as a counter for the current nesting level.

(Of course, "resource-friendly" is ignoring the fact that this interpreter was written in Java -- I just wanted to write some quick code and Java just happened to be what I wrote it in.)

Each time you find a '[', push the current position (or another "marker" token or a "context") on a stack. When you come accross a ']', you're at the end of the loop, and you can pop the marker token from the stack.

Since in BF the '[' already checks for a condition and may need jump past the ']', you may want to have a flag indicating that instructions shall be skipped in the current loop context.

Python 3.0 example of the stack algorithm described by the other posters:

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()))))

(Well, to be honest, this only finds matching brackets. I don't know brainf*ck, so what to do next, I have no idea.)

And here's the same code I gave as an example earlier in C++, but ported to VB.NET. I decided to post it here since Gary mentioned he was trying to write his parser in a BASIC dialect.

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

I don't have sample code, but.

I might try using a stack, along with an algorithm like this:

  1. (executing instruction stream)
  2. Encounter a [
  3. If the pointer == 0, then keep reading until you encounter the ']', and don't execute any instructions until you reach it.. Goto step 1.
  4. If the pointer !=0, then push that position onto a stack.
  5. Continue executing instructions
  6. If you encounter a ]
  7. If pointer==0, pop the [ off of the stack, and proceed (goto step 1)
  8. If pointer != 0, peek at the top of the stack, and go to that position. (goto step 5)

This question is a bit old, but I wanted to say that the answers here helped me decide the route to take when writing my own Brainf**k interpreter. Here's the final product:

#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);
        }
    }
}
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;
    }

}

This should work. You need to modify it somewhat. That is actually standard example how brainfuck interpreter works. I modified it to use in my app, brackets are handled there:

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;
}

It looks like this question has become a "post your bf interpreter" poll.

So here's mine that I just got working:

#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;
}
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