我最终创建了Brainfuck解析器(使用BASIC方言)来创建解释器,但我意识到它不像我最初想象的那样简单。我的问题是我需要一种方法来准确解析Brainfuck程序中的匹配循环运算符。这是一个示例程序:
,>,>++++++++[<------<------>>-]
<<[>[>+>+<<-]>>[<<+>>-]<<<-]
>>>++++++[<++++++++>-],<.>.
'['=循环开始
']'=循环结束
我需要记录每个匹配循环运算符的起点和终点,以便我可以根据需要在源代码周围跳转。有些循环是单独的,有些则是嵌套的。
解析此问题的最佳方法是什么?我在想,也许在源文件中移动以创建一个2D数组(或类似的记录),以记录每个匹配运算符的开始和结束位置,但这似乎在源中进行了很多“往返”操作。这是最好的方法吗?
更多信息:Brainfuck homepage
编辑:非常感谢任何语言的示例代码。
最佳答案
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“[”时,都将指令指针的当前位置压入该堆栈。每当遇到“]”时,都将指令指针重置为堆栈顶部当前的值。循环完成后,将其弹出堆栈。
这是一个具有100个存储单元的C++示例。该代码以递归方式处理嵌套循环,尽管未进行完善,但应说明这些概念。
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;
}
编辑
我只是再次浏览了一下代码,我意识到while循环中存在一个错误,如果指针的值为0,它将“跳过”已解析的循环。这是我进行更改的地方:
while((tmp >= (sp + 1)) || *ip != ']') // the bug was tmp > (sp + 1)
{
ip++;
if(*ip == '[')
stack[++tmp] = ip - program;
else if(*ip == ']')
tmp--;
}
以下是相同解析器的实现,但未使用递归:
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;
}
关于parsing - 创建Brainfuck解析器,解析循环运算符的最佳方法是什么?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1055758/