预测分析程序.cpp

编译原理预测分析实验

#include "stdio.h"
int flag=1;
int ZT=1;
int k=0,j=0,l=0;
char Stack[10]={'\0'};
char S[80]={'\0'};
char T[100]={'\0'};
char top,coda,tp;
char bring[20][20]={"error\n","E->TA\n","A->+TA\n","A->-TA\n","T->PB\n","B->*PB\n","B->/PB\n","P->c\n",
"P->i\n","A->empty\n","B->empty\n","\n"};
void Push()
{
   Stack[k]=tp;top=tp; k++;
}
void Pop()
{
   k--; Stack[k]='\0';top=Stack[k-1];
}
void Next()
{
coda=S[j];j++;
}
void g()
{
int i=0;
while(S[i]!='\0'&&S[i]!=' ')
   i++;
S[i]='#';
}
void copy(int n)
{
   int i;
   for(i=0;i<10;i++,l++)
    T[l]=bring[n][i];
   if(n==0) ZT=0;
}
void emptyST()
{int i;
	for(i=0;i<100;i++)
   {
	S[i]='\0';T[i]='\0';
	}
}
void display(int n)
{
	int i;
	for(l=0;l<k;l++) T[l]=Stack[l];
	while(l<10) {T[l]=' ';l++;}
	for(i=0;i<j-1;i++,l++) 
		T[l]=' ';
	for(;S[i]!='\0';i++,l++)   
		T[l]=S[i];
	while(l<30) {T[l]=' ';l++;}
	copy(n);
	printf("%s",T);

}
void match(char a,char b)
{	
   if(a==b)
     if(a=='#'){flag=0;return;}
     else
       {Pop();Next();display(11);flag=1;return;}
   else	   
   switch(a)
   {	   	   	
		case 'E':
      switch(b)
        {case 'c':
         case 'i':
            {    			
				Pop();
                tp='A';Push();
                tp='T';Push();    /* E-> TA */
				
                display(1);
                break;
             }
         case '#':{Pop();display(11);break;}
         default:{Next();display(0);}     /* error */
         }
     break;
    case 'A':
        switch(b)
        {case'+':
            {    tp='T';Push();
                 tp='+';Push();      /* A-> +TA */
                 display(2);
                 break;
             }
         case '-':
            {    tp='T';Push();
                tp='-';Push();    /* A-> -TA */
                display(3);
                break;
             }
         case '#':{Pop();display(9);break;}       /*   A->empty */
         default:{Next();display(0);}           /*   error   */
         }
     break;
    case 'T':
        switch(b)
        {case 'c':
         case 'i':
            {    Pop();
                tp='B';Push();
         tp='P';Push();      /* T-> PB */
                display(4);
                break;
             }
         case '+':
         case '-':
         case '#': {Pop();display(0);break;} /*   error   */
         default:{Next();display(0);}   /*   error   */
         }
     break;
    case 'B':
         switch(b)
        {case '*':
      {     tp='P';Push();
			tp='*';Push();      /* B-> *PB */
                  display(5);
                  break;
              }
         case '/':
      {    tp='P';Push();
			tp='/';Push();    /* B-> /PB */
                 display(6);break;
              }
         case '+':
         case '-':
         case '#':{Pop();display(10);break;}        /* B->empty */
         default:{Next();display(0);}               /*   error   */
         }
     break;
    case 'P':
         switch(b)
        {case 'c':
      {    Pop();
			tp='c';Push();             /* P-> c */
			display(7);break;
              }
         case 'i':
      {     Pop();
			tp='i';Push();            /* P-> i */
                  display(8);break;
              }
         default:{Pop();display(0);}       /* error   */
         }
    }
    flag=1;return;
}
main()
{
   char i='1';      
   int w=0;
   char buf[10][12]={"E->TA","A->+TA","A->-TA","A->e","T->PB","B->*PB","B->/PB","B->e","P->c","P->i"};
   printf("以下为已知文法G[E]:\n");
   for (;w<10;w++)
   {
    printf("%s     ",buf[w]);
    if ((w+1)%4==0) printf("\n");
    }
   printf("\n\n\n其中：\n\nE,T,A,P,B为非终结符\ni,c,e为终结符\n+,-,*,/为字符\n\n");
   printf("请输入要预测的字符串:\n");
   scanf("%s",S);
   while(S[0]!='#') 
   {
   k=0;j=0;flag=1;
   tp='#';Push();
   tp='E';Push();
   g();
   Next();
   display(11);
   while(flag)
      {
        match(top,coda);
       }
   if(ZT)
   {
     printf("该输入串是文法G[E]的句子!\n");
    }
   else
    {
     printf("该输入串不是文法G[E]的句子!\n");
     }
   emptyST(); i++;ZT=1;
   printf("请输入要预测的字符串:\n");
   scanf("%s",S);
   if(S[0]=='#')
	 //  exit(0);
	 return 0;
   }
}