实验二——递归下降语法分析

  1 #include<stdio.h>
  2 #include<stdlib.h>
  3 #include<string.h>
  4 char prog[800],dc[8];
  5 char ch;
  6 int syn,p,m=0;
  7 int n,sum=0;
  8 char *word[6]={"begin","if","then","while","do","end"};
  9 int kk=0;
 10 void scaner();
 11 void lrparser();
 12 void yucu();
 13 void statement();
 14 void condition();
 15 void expression();
 16 void term();
 17 void factor();
 18 void scaner(){
 19     m=0;
 20     for(n=0;n<8;n++){
 21         dc[n]=NULL;
 22     }
 23     ch=prog[p++];
 24     while(ch==‘ ‘){
 25         ch=prog[p];
 26         p++;
 27     }
 28     //标识符 
 29     if((ch>=‘a‘&&ch<=‘z‘)||(ch>=‘A‘&&ch<=‘Z‘)){
 30         m=0;
 31         while((ch>=‘0‘&&ch<=‘9‘)||(ch>=‘a‘&&ch<=‘z‘)||(ch>=‘A‘&&ch<=‘Z‘)){
 32             dc[m++]=ch;
 33             ch=prog[p++];
 34         }
 35         p--;
 36         syn=10;
 37         //保留字 
 38         for(n=0;n<6;n++){
 39             if(strcmp(dc,word[n])==0){
 40                 syn=n+1;
 41                 break;
 42             }
 43         }
 44     }
 45     //数字 
 46     else if((ch>=‘0‘&&ch<=‘9‘)){
 47         sum=0;
 48         while((ch>=‘0‘&&ch<=‘9‘)){
 49             sum=sum*10+ch-‘0‘;
 50             ch=prog[p++];
 51         }
 52         p--;
 53         syn=11;
 54     }
 55     else{
 56         switch(ch){
 57             case ‘<‘:m=0;
 58                 dc[m++]=ch;
 59                 ch=prog[p++];
 60                 if(ch==‘>‘){
 61                     syn=22;
 62                     dc[m++]=ch;
 63                 }
 64                 else if(ch==‘=‘){
 65                     syn=21;
 66                     dc[m++]=ch;
 67                 }
 68                 else{
 69                     syn=20;
 70                     p--;
 71                 }
 72                 break;
 73             case ‘>‘:m=0;
 74                 dc[m++]=ch;
 75                 ch=prog[p++];
 76                 if(ch==‘=‘){
 77                     syn=24;
 78                     dc[m++]=ch;
 79                 }
 80                 else{
 81                     syn=23;
 82                     p--;
 83                 }
 84                 break;
 85             case ‘:‘:m=0;
 86                 dc[m++]=ch;
 87                 ch=prog[p++];
 88                 if(ch==‘=‘){
 89                     syn=18;
 90                     dc[m++]=ch;
 91                 }
 92                 else{
 93                     syn=17;
 94                     p--;
 95                 }
 96                 break;
 97             case ‘*‘:syn=15;dc[0]=ch;break;
 98             case ‘/‘:syn=16;dc[0]=ch;break;
 99             case ‘+‘:syn=13;dc[0]=ch;break;
100             case ‘-‘:syn=14;dc[0]=ch;break;
101             case ‘=‘:syn=25;dc[0]=ch;break;
102             case ‘;‘:syn=26;dc[0]=ch;break;
103             case ‘(‘:syn=27;dc[0]=ch;break;
104             case ‘)‘:syn=28;dc[0]=ch;break;
105             case ‘#‘:syn=0;dc[0]=ch;break;
106             case ‘\n‘:syn=-2;dc[0]=ch;break;        
107         }
108     }
109 }
110 
111 void lrparser(){
112     if (syn==1) {
113         scaner();
114         yucu();
115         if (syn==6) {
116             scaner();
117             if (syn==0 && kk==0){
118                 printf("success \n");
119             } 
120         } 
121         else {
122             printf("error,lose ‘end‘ ! \n");
123             kk=1;
124             exit(0);
125         }
126     } 
127     else {
128         printf("error,lose ‘begin‘ ! \n");
129         kk=1;
130         exit(0);
131     }
132     return;
133 }
134 
135 void yucu() {
136     statement();
137     while(syn==26) { 
138         scaner();
139         statement();
140     }
141     return;
142 }
143 
144 void statement(){
145     if (syn==10) {
146         scaner();
147         if (syn==18) {
148             scaner();
149             expression();
150         } 
151         else {
152             printf("error!");
153             kk=1;
154             exit(0);
155         }
156     }
157     else if(syn==2){
158         condition();
159         scaner();
160         if(syn==3){
161             statement(); 
162         }
163         else{
164             printf("error,lose ‘then‘ ! \n");
165             kk=1;
166             exit(0);
167         }
168     }
169     else {
170         printf("error!");
171         kk=1;
172         exit(0);
173     }
174     
175     return;
176 }
177 
178 void condition(){
179     expression();
180     if(syn==25||syn==0||syn==20||syn==21||syn==23||syn==24){
181         scaner(); 
182     }
183     else{
184         printf("error! \n");
185         kk=1;
186         exit(0);
187     }
188     expression();
189     return;
190 } 
191 
192 void expression(){
193     term();
194     while(syn==13 || syn==14) {
195         scaner();
196         term();
197     }
198     return;
199 }
200 
201 void term(){
202     factor();
203     while(syn==15 || syn==16) {
204         scaner();
205         factor();
206     }
207     return;
208 }
209 
210 void factor(){
211     if(syn==10 || syn==11){
212         scaner();
213     }
214     else if(syn==27) {
215         scaner();
216         expression();
217         if(syn==28){
218             scaner();
219         }
220         else {
221             printf(" ‘)‘ 错误\n");
222             kk=1;
223             exit(0);
224         }
225     } else {
226         printf("表达式错误\n");
227         kk=1;
228         exit(0);
229     }
230     return;
231 }
232 int main(void){
233     p=0;
234     int i;
235     printf("请输入:\n");
236     do {
237         scanf("%c",&ch);
238         prog[p++]=ch;
239     } while(ch!=‘#‘);
240     p=0;
241     do{
242         scaner();
243         lrparser();
244     }while(syn!=0);
245     printf("语法分析结束！\n");
246     
247 }