问题
I'm creating a Brainfuck parser (in a BASIC dialect) ultimately to create an interpreter but i've realise it's not as straight forward as i first thought. My problem is that i need a way to accurately parse the matching loop operators within a Brainfuck program. This is an example program:
,>,>++++++++[<------<------>>-]
<<[>[>+>+<<-]>>[<<+>>-]<<<-]
>>>++++++[<++++++++>-],<.>.
'[' = start of loop
']' = end of loop
I need to record the start and end point of each matching loop operator so i can jump around the source as needed. Some loops are alone, some are nested.
What would be the best way to parse this? I was thinking maybe move through the source file creating a 2D array (or such like) recording the start and end positions of each matching operator, but this seems like a lot of 'to'ing and fro'ing' through the source. Is this the best way to do it?
More info: Brainfuck homepage
EDIT: Sample code in any language greatly appreciated.
回答1:
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;
}
回答2:
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.
回答3:
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.nestingof0means 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:
If an
[is encountered while seeking for the corresponding loop closing], then thenestingvariable is incremented by1in order to indicate that we have entered a nested loop.If an
]is encountered, and:a. If the
nestingvariable is greater than0, then thenestingvariable is decremented by1to indicate that we've left a nested loop.b. If the
nestingvariable is0, then we know that the end of the loop has been encountered, so seeking the end of the loop in thewhileloop is terminated by executing abreakstatement.
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.)
回答4:
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.
回答5:
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.)
回答6:
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
回答7:
I don't have sample code, but.
I might try using a stack, along with an algorithm like this:
- (executing instruction stream)
- Encounter a [
- If the pointer == 0, then keep reading until you encounter the ']', and don't execute any instructions until you reach it.. Goto step 1.
- If the pointer !=0, then push that position onto a stack.
- Continue executing instructions
- If you encounter a ]
- If pointer==0, pop the [ off of the stack, and proceed (goto step 1)
- If pointer != 0, peek at the top of the stack, and go to that position. (goto step 5)
回答8:
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);
}
}
}
回答9:
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;
}
回答10:
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;
}
来源：https://stackoverflow.com/questions/1055758/creating-a-brainfuck-parser-whats-the-best-method-of-parsing-loop-operators