算术表达式的语法分析器

(华保健老师)在这个题目中，要求你完成一个针对算术表达式的语法分析器。该算术表达式的上下文无关文法是：

E -> E + T

   | E - T

   | T

T -> T * F

   | T / F

   | F

F -> num

   | (E)

#include <ctype.h>#include <stdio.h>#include <stdlib.h>void parse_F();void parse_T();void parse_E();void error (char *want, char got);int i;char *str = 0;void error (char *want, char got){  fprintf (stderr, "Compling this expression:\n%s\n", str);  int j = i;  while (j--)    fprintf (stderr, " ");  fprintf (stderr, "^\n");  fprintf (stderr, "Syntax error at position: %d\n"           "\texpecting: %s\n"           "\tbut got  : %c\n",           i, want, got);  exit (0);  return;}void parse_F(){  char c = str[i];  if (isdigit(c)){    i++;    return;  }  if (c=='('){    i++;    parse_E();    c = str[i];    if (c==')'){      i++;      return;    }    error ("\')\'", c);    return;  }  error ("\'0-9\' or \'(\'", c);  return;}void parse_T(){  parse_F();  char c = str[i];  while (c=='*' || c== '/'){    i++;    parse_F();    c = str[i];  }  return;}void parse_E(){  parse_T();  char c = str[i];  while (c=='+' || c=='-'){    i++;    parse_T();    c = str[i];  }  return;}void parse (char *e){  str = e;  i = 0;  parse_E();  if (str[i]=='\0')    return;  error ("\'+\' or '\\0\'", str[i]);  return;}///////////////////////////////////////////////// Your job:// Add some code into the function parse_E() and// parse_T to parse "-" and "/" correctly.// When you finish your task, NO error message// should be generated.// Enjoy! :-Pint main (char argc, char **argv){  // There are the following rules on an expression:  //   1. Every expression is represented as a string;  //   2. integers are non-negative;  //   3. integers are between 0-9.  char *e;  e = "(2)";  parse(e);  e = "(3+4*5)";  parse(e);  e = "(8-2)*3";  parse(e);  e = "(8-2)/3";  parse(e);    return 0;}