我是C的新手。目前我正在尝试写一个Brainfuck解释器。到目前为止我已经尝试过了。
#include <unistd.h>
#include <stdlib.h>
char *line;
int curr_pos;
void interprete(char *coms)
{
int a;
int curr_loop;
a = -1;
curr_loop = 0;
while (line[++a])
line[a] = 0;
a = -1;
while (coms[++a])
{
if (coms[a] == '+')
line[curr_pos]++;
else if (coms[a] == '-')
line[curr_pos]--;
else if (coms[a] == '>')
curr_pos++;
else if (coms[a] == '<')
curr_pos--;
else if (coms[a] == '.')
write(1, &line[curr_pos], 1);
else if (coms[a] == '[')
{
if (line[curr_pos])
curr_pos++;
else
{
curr_loop = 1;
while (curr_loop)
{
++a;
if (coms[a] == '[')
curr_loop++;
else if (coms[a] == ']')
curr_loop--;
}
}
}
else if (coms[a] == ']')
{
if (line[curr_pos])
{
curr_loop = 1;
while (curr_loop)
{
--a;
if (coms[a] == '[')
curr_loop--;
else if (coms[a] == ']')
curr_loop++;
}
}
else
curr_pos++;
}
}
}
int main(int ac, char **av)
{
if (ac == 2)
{
curr_pos = 0;
line = malloc(sizeof(char) * 4096);
interprete(av[1]);
}
write(1, "\n", 1);
}
它只能在没有循环的情况下工作(&#34; [&#34;和&#34;]&#34;)。当我尝试时:
++++++++++[>+++++++>++++++++++>+++>+<<<<-]>++.>+.+++++++..+++.>++.<<+++++++++++++++.>.+++.------.--------.>+.>.
它给我输出
^B^A^H^H^K^B^Q^K^N^H^@^C^@
预期产出:
Hello World!
答案 0 :(得分:1)
我猜问题出现在以下代码块中:
else if (coms[a] == '[')
{
...
}
else if (coms[a] == ']')
{
...
}
程序正在寻找另一个括号(并找到它),但你的代码指针在while
- 语句中增加了(第17行)。因此,在搜索循环后,您必须按a
递减1
。第二个问题是,如果curr_pos
和com[a] == '['
以及{line[curr_pos] != 0
,请增加数据 -pointer(com[a] == ']'
)。 1}}和line[curr_pos] == 0
您必须递增代码 -pointer(a
)仍在while
- 语句中自动递增的内容。所以实际上你必须通过这种情况。最后,如果当前单元格值不等于零,您实际上不必检查两个括号。我建议的代码如下所示:
else if (coms[a] == '[')
{
if (!line[curr_pos])
{
curr_loop = 1;
while (curr_loop)
{
++a;
if (coms[a] == '[')
cur_loop++;
else if (coms[a] == ']')
cur_loop--;
}
a--;
}
}
else if (coms[a] == ']')
{
// You can always jump back to the opening bracket '['
// because then the program checks again and jumps behind
// the closing bracket if line[a] != 0
curr_loop = 1;
while (curr_loop)
{
--a;
if (coms[a] == '[')
curr_loop--;
else if (coms[a] == ']')
curr_loop++;
}
a--;
}
顺便说一下:尝试实现',
' - 命令。它使得brainfuck程序更有趣;)