词法分析程序的设计与实现

  1 #include<stdio.h>
  2 #include<conio.h>
  3 #include<math.h>
  4 #include<string.h>
  5 #include<stdlib.h>
  6 
  7 int i, row = 0, line = 0;
  8 char test[1000];  //test文件中的字符 
  9 int number[100];    //常数表 
 10 char mark[100][5];   //标识符表
 11 
 12 //词法分析
 13 int wordanalysis()
 14 {
 15     //标识符和保留字
 16     if ((test[i] >= ‘A‘&&test[i] <= ‘Z‘)||(test[i]>=‘a‘&&test[i]<=‘z‘))  
 17     {
 18         char word[10];
 19         //保留字表
 20         char pro[100][100] = { "PROGRAM", "BEGIN", "END", "VAR", "INTEGER", "WHILE",
 21                                  "IF", "THEN", "ELSE", "DO", "PROCEDURE" ,"char",
 22                                 "int","if","else","var" ,"return","break",
 23                                 "do","while","for","double","float","short"}; 
 24 
 25         int n = 0;
 26         word[n++] = test[i++];
 27         while ((test[i] >= ‘A‘&&test[i] <= ‘Z‘) || (test[i] >= ‘0‘ && test[i] <= ‘9‘)||(test[i]>=‘a‘&&test[i]<=‘z‘))
 28         {
 29             word[n++] = test[i++];
 30         }
 31         word[n] = ‘\0‘;
 32         i--;
 33 
 34         //判断该标识符是否为保留字
 35         for (n = 0; n < 100; n++)
 36         {
 37             if (strcmp(word, pro[n]) == 0)
 38             {
 39                 printf(">> %s\t(%d,-) 保留字\n", pro[n], n + 1);
 40                 return 3;
 41             }
 42         }
 43 
 44         //判断该标识符是否在标识符表中
 45         int m = 0;
 46         if (line != 0)
 47         {
 48             int q = 0;
 49             while (q<line)
 50             {
 51                 if (strcmp(word, mark[q++]) == 0)
 52                 {
 53                     printf(">> %s\t(10,%d) 标识符\n", word, q);
 54                     return 3;
 55                 }
 56             }
 57 
 58         }
 59 
 60         //将该标识符保存到标识符表中
 61         strcpy(mark[line], word);
 62         
 63         printf(">> %s\t(10, %d) 标识符\n", word, line + 1);
 64         line++;
 65         return 3;
 66 
 67     }
 68     //数字 
 69     else if (test[i] >= ‘0‘ && test[i] <= ‘9‘)  
 70     {
 71         char x[100];
 72         int n = 0;
 73         x[n++] = test[i++];
 74         
 75         while (test[i] >= ‘0‘ && test[i] <= ‘9‘)
 76         {
 77             x[n++] = test[i++];
 78         }
 79         x[n] = ‘\0‘;
 80         i--;
 81         int num = atoi(x); //将字符串转换成int型
 82         
 83         //判断该常数是否存在于常数表中
 84         if (row != 0)
 85         {   
 86             
 87             for(int y=0;y<row;y++)
 88             {
 89                 if(number[y]==num)
 90                 {
 91                     printf(">> %d\t(11,%d)\n", num, y + 1);
 92                     return 3;
 93                 }
 94             }
 95         }
 96         
 97         //将该常数保存到标识符表中
 98         number[row]=num;
 99         
100 
101         int line = row;
102         printf(">> %d\t(11,%d)\n", num, line + 1);
103         row++;
104 
105         return 3;
106     }
107     
108     //各种符号
109     else                      
110         switch (test[i])
111     {
112         case ‘ ‘:
113         case ‘\n‘:
114             return -1;
115         case ‘#‘: return 0;
116         case ‘=‘:printf(">> =\t(25,-)\n"); return 3;
117         case ‘<‘:
118             i++;
119             if (test[i] == ‘=‘)
120             {
121                 printf(">> <= \t(21,-)\n");
122                 return 3;
123             }
124             else if (test[i] == ‘>‘)
125             {
126                 printf(">> <>\t(22,-)\n");
127                 return 3;
128             }
129             else
130             {
131                 i--;
132                 printf(">> <\t(20,-)\n");
133                 return 3;
134             }
135         case ‘>‘:
136             i++;
137             if (test[i] == ‘=‘)
138             {
139                 printf(">> >=\t(24,-)\n");
140                 return 3;
141             }
142             else
143             {
144                 i--;
145                 printf(">> >\t(23,-)\n");
146                 return 3;
147             }
148         case ‘+‘: printf(">> +\t(13,-)\n"); return 3;
149         case ‘-‘: printf(">> -\t(14,-)\n"); return 3;
150         case ‘*‘: printf(">> *\t(15,-)\n"); return 3;
151         case ‘/‘: 
152             i++;
153             if(test[i]!=‘/‘){
154                 i--;
155                 printf(">> /\t(16,-)\n"); return 3;
156             }
157 
158             else{
159 
160                 while(1){
161                     if(test[i++]==‘\n‘)
162                         return -1;
163                 }
164                 printf(">> //\t(37,-)\n");return 3;
165 
166             }
167 
168         case ‘:‘: printf(">> :\t(17,-)\n"); return 3;
169         case ‘;‘: printf(">> ;\t(26,-)\n"); return 3;
170         case ‘(‘: printf(">> (\t(27,-)\n"); return 3;
171         case ‘)‘: printf(">> )\t(28,-)\n"); return 3;
172 
173     }
174 
175 }
176 
177 int main()
178 {
179 
180     int c = 0;
181     int m;
182     i = 0;
183     FILE *fp;
184     fp=fopen(".\\test.txt","r");
185     if (fp == NULL)
186     {
187         printf("can‘t open file!\n");
188         exit(0);
189     }
190 
191     while (!feof(fp))
192     {
193         test[c++] = fgetc(fp);
194     }
195     test[c] = ‘#‘;
196     do
197     {
198         m = wordanalysis();
199 
200         switch (m)
201         {
202         case -1:i++; break;
203         case 0: i++; break;
204         case 3: i++; break;
205         }
206     } while (m != 0);
207 
208 
209     return 0;
210 }