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)
部分代码来自MOOC
#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;}与君共勉
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