• CSUFT 编译原理实验四 LR(1)


      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 #include <stack>
      5 #include <queue>
      6 #include <map>
      7 #include <algorithm>
      8 #include <vector>
      9 
     10 using namespace std;
     11 
     12 char action[10][3][5]= {{"S3#","S4#","NUll"},
     13                       {"NUll","NUll","acc"},
     14                       {"S6#","S7#","NUll"},
     15                       {"S3#","S4#","NUll"},
     16                       {"r3#","r3#","NUll"},
     17                       {"NUll","NUll","r1#"},
     18                       {"S6#","S7#","NUll"},
     19                       {"NUll","NUll","r3#"},
     20                       {"r2#","r2#","NUll"},
     21                       {"NUll","NUll","r2#"}
     22                      };
     23 /* GOTO表*/
     24 int goto1[10][2]= {1,2,
     25                    0,0,
     26                    0,5,
     27                    0,8,
     28                    0,0,
     29                    0,0,
     30                    0,9,
     31                    0,0,
     32                    0,0,
     33                    0,0
     34                   };
     35 char vt[3]= {'a','b','#'}; /*存放终结符*/
     36 char vn[2]= {'S','B'};  /*存放非终结符*/
     37 char *LR[5]= {"E->S#","S->BB#","B->aB#","B->b#"}; /*存放产生式*/
     38 /*输出状态栈、输出符号栈、输出输入串*/
     39 
     40 int main()
     41 {
     42    // printf("%s
    
    " ,action[0][1]);
     43     int y,z,m,n,top,top1,top2,top3,j,k,i,g,h,l,p;
     44     char x;
     45     char b[10];//符号栈
     46     char str[10];//输入串
     47     int a[10];//状态栈
     48     char copy1[20],copy[20];
     49     int count = 0;
     50     top = top1 = top2 = top3 = 0;
     51       memset(a,0,sizeof(a));
     52     a[top1]=0;
     53 
     54     b[top2] = '#';
     55 
     56     scanf("%s",str);
     57     z=0;
     58     top3 =  strlen(str);
     59     int flag = 0;
     60     do
     61     {
     62         y=z;
     63         //if() break;
     64         m=0;
     65         n=0;              /*y,z指向状态栈栈顶*/
     66         g=top;
     67         j=0;
     68         k=0;
     69         x=str[top];
     70         count++;
     71         printf("%d	",count);
     72         while(m<=top1)                /*输出状态栈*/
     73         {
     74             printf("%d",a[m]);
     75             m=m+1;
     76         }
     77         printf("		");
     78         while(n<=top2)                  /*输出符号栈*/
     79         {
     80             printf("%c",b[n]);
     81             n=n+1;
     82         }
     83         printf("		");
     84         while(g<top3)                 /*输出输入串*/
     85         {
     86             printf("%c",str[g]);
     87             g=g+1;
     88         }
     89         printf("		");
     90 
     91 
     92 
     93     /*查动作表*/
     94     if(x == 'a')j=0;
     95     if(x == 'b')j=1;
     96     if(x == '#')j=2;
     97     if(action[y][j]=="NUll"){
     98         printf("error
    ");
     99          return 0;
    100     }
    101 
    102     strcpy(copy,action[y][j]);
    103 
    104     /*处理移进*/
    105     if(copy[0]=='S')
    106     {
    107         z=copy[1]-'0';
    108         top1=top1+1;
    109         top2=top2+1;
    110         a[top1]=z;
    111         b[top2]=x;
    112         top=top+1;
    113         i=0;
    114         while(copy[i]!='#')
    115         {
    116             printf("%c",copy[i]);
    117             i++;
    118         }
    119         printf("
    ");
    120     }
    121 
    122     /*处理归约*/
    123     if(copy[0]=='r')
    124     {
    125         i=0;
    126         while(copy[i]!='#')
    127         {
    128             printf("%c",copy[i]);
    129             i++;
    130         }
    131         h=copy[1]-'0';
    132         strcpy(copy1,LR[h]);
    133         if(copy1[0]=='S')k=0;
    134         if(copy1[0]=='B')k=1;
    135         l=strlen(LR[h])-4;
    136         top1=top1-l+1;
    137         top2=top2-l+1;
    138         y=a[top1-1];
    139         p=goto1[y][k];
    140         a[top1]=p;
    141         b[top2]=copy1[0];
    142         z=p;
    143         printf("		");
    144         printf("%d
    ",p);
    145     }
    146     if(copy[0]=='a') { printf("%s
    ",copy); flag =1;}
    147 
    148     if(copy[0]=='N') {printf("error
    ");flag = 1;}
    149     if(flag) break;
    150     }while(1);
    151     return 0;
    152 }
    View Code

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  • 原文地址:https://www.cnblogs.com/lmlyzxiao/p/5626681.html
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