读龙书学编译原理 语法分析(6)...

这两天都没更, 主要是马上要连着要考三门(16, 18, 20), 都没时间学编译器了, 等过了这段时间应该能有很多时间来搞这个, 之后我准备先把之前写的那个词法分析器生成器改造一下, 弄个最简版的正则引擎出来玩玩, 不过这些都是后话了, 今天似乎也没什么时间, 也就把这单元的作业写了一下... 代码比较简单就不多做解释了...

  1 #include <ctype.h>
  2 #include <stdio.h>
  3 #include <stdlib.h>
  4
  5 void parse_F();
  6 void parse_T();
  7 void parse_E();
  8 void error (char *want, char got);
  9
 10 int i;
 11 char *str = 0;
 12
 13 void error (char *want, char got)
 14 {
 15     fprintf (stderr, "Compling this expression:\n%s\n", str);
 16     int j = i;
 17     while (j--)
 18         fprintf (stderr, " ");
 19     fprintf (stderr, "^\n");
 20     fprintf (stderr, "Syntax error at position: %d\n"
 21             "\texpecting: %s\n"
 22             "\tbut got  : %c\n",
 23             i, want, got);
 24     exit (0);
 25     return;
 26 }
 27
 28 void parse_F()
 29 {
 30     char c = str[i];
 31     if (isdigit(c)){
 32         i++;
 33         return;
 34     }
 35     if (c==‘(‘){
 36         i++;
 37         parse_E();
 38         c = str[i];
 39         if (c==‘)‘){
 40             i++;
 41             return;
 42         }
 43         error ("\‘)\‘", c);
 44         return;
 45     }
 46     error ("\‘0-9\‘ or \‘(\‘", c);
 47     return;
 48 }
 49
 50
 51 void parse_T()
 52 {
 53     parse_F();
 54     char c = str[i];
 55     while (c==‘*‘ || c == ‘/‘){
 56         i++;
 57         parse_F();
 58         c = str[i];
 59     }
 60     return;
 61 }
 62
 63 void parse_E()
 64 {
 65     parse_T();
 66     char c = str[i];
 67     while (c ==‘+‘ || c == ‘-‘){
 68         i++;
 69         parse_T();
 70         c = str[i];
 71     }
 72     return;
 73 }
 74
 75 void parse (char *e)
 76 {
 77     str = e;
 78     i = 0;
 79     parse_E();
 80     if (str[i]==‘\0‘)
 81         return;
 82     error ("\‘+\‘ or ‘\\0\‘", str[i]);
 83     return;
 84 }
 85 ///////////////////////////////////////////////
 86 // Your job:
 87 // Add some code into the function parse_E() and
 88 // parse_T to parse "-" and "/" correctly.
 89 // When you finish your task, NO error message
 90 // should be generated.
 91 // Enjoy! :-P
 92 int main (int argc, char **argv)
 93 {
 94     // There are the following rules on an expression:
 95     //   1. Every expression is represented as a string;
 96     //   2. integers are non-negative;
 97     //   3. integers are between 0-9.
 98     char *e;
 99
100     e = "(2)";
101     parse(e);
102
103     e = "(3+4*5))";
104     parse(e);
105
106     e = "(8-2)*3";
107     parse(e);
108
109     e = "(8-2)/3";
110     parse(e);
111
112     return 0;
113 }