关于编译中LL1文法的判别,拜求编译高手指点!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

最近小弟在学编译,要做一个LL1文法的识别程序,小弟想用C或C++来做,初步设想是文法输入形式为S->AB A->aA,A->b,B->cB,B->s,然后程序判断是否为LL1.

想请教各位高手预测分析表怎么做?还有关于文法的机内存储形式的相关操作.

8 个解决方案

#1

程序名称： LL(1)文法分析程序
http://www.pudn.com/downloads56/sourcecode/chinese/detail199479.html

#2

LL1 文法分析的实现
http://www.yuanma.org/data/2006/0720/article_1174.htm

#3

LL(1)文法及分析表的自动构造
http://scholar.ilib.cn/Abstract.aspx?A=shdegydxxb200102007

#4

LL1算法
点击数：229 发布日期：2006-5-12 18:15:00
http://www.programfan.com/blog/article.asp?id=13961

#5

真心的谢谢两位!
不过这些网页怎么都打不开?

#6

to 楼主：
http://www.programfan.com/blog/article.asp?id=13961

这个能打开～

#7

//文法是：E->TE' E'->+TE' T->FT' T'->+ET' F->(E)/i
#include <iostream.h>
#include <stdlib.h>
#include <string.h>
#include<stdio.h>
typedef char datatype;
#define maxsize 100
int save_i=0;
int save_j=0;
int l=0;
int w;
char a[10];
typedef struct
{
datatype data[maxsize];
int top;
}seqstack;
seqstack g;
void Input()//输入流
{
  int i=-1;
char c;
    c=getchar();
     while(c!='#')
{
  a[++i]=c;

  c=getchar();

}
    a[++i]='$';
  save_i=i;
}
void SETNULL(seqstack *s)
{
s->top=-1;
}
int EMPTY(seqstack *s)
{

if(s->top>=0) return 0;
    else return 1;
}
int PUSH(seqstack *s,datatype x)
{
if(s->top==maxsize-1){printf("overflow");}
else
{
  s->top++;
  s->data[s->top]=x;
  l=s->top;
}
return l;
}
int  POP(seqstack *s)
{
if(EMPTY(s))
{printf("underflow"); return NULL;}
else{
  s->top--;
  l=s->top;
  //return (s->data[s->top+1]);

}
  return l;
}
datatype TOP(seqstack *s)
{
if(EMPTY(s))
{printf("stack is empty ");
return NULL;} //空栈
else
return (s->data[s->top]);
}
int LL1()
{
char c;
if(g.data[g.top]==a[save_j])
{save_j++;
POP(&g);
    return save_j;}
    else if(a[save_j]=='i')           ////i--id;t--T';e--E'
{
  c=TOP(&g);
  if(c=='E')
  {
   POP(&g);
   PUSH(&g,'e');
   PUSH(&g,'T');
  }
  if(c=='T')
  {POP(&g);
   PUSH(&g,'t');
   PUSH(&g,'F');
  }
   if(c=='F')
   {POP(&g);
   PUSH(&g,'i');

   }
}
else if(a[save_j]=='+')
{
  c=TOP(&g);
  if(c=='e')
  {
   POP(&g);
   PUSH(&g,'e');
   PUSH(&g,'T');
   PUSH(&g,'+');

  }
  if(c=='t')
  {
   POP(&g);
  }
}
else if(a[save_j]=='*')
{
   c=TOP(&g);
   if(c=='t')
   { POP(&g);
   PUSH(&g,'t');
   PUSH(&g,'F');
   PUSH(&g,'*');

   }
}
else if(a[save_j]=='(')
{
   c=TOP(&g);
   if(c=='E')
   {POP(&g);
       PUSH(&g,'e');
      PUSH(&g,'T');
   }
   if(c=='T')
   { POP(&g);
   PUSH(&g,'t');
   PUSH(&g,'F');
  /// PUSH(&g,'*');

   }
   if(c=='F')
   { POP(&g);
   PUSH(&g,')');
   PUSH(&g,'E');
   PUSH(&g,'(');
   }
}
else if(a[save_j]==')')
{
         c=TOP(&g);
   if(c=='e')
   {
    POP(&g);
      }
   if(c=='t')
   {
    POP(&g);
   }

}
else if(a[save_j]=='$')
{
        c=TOP(&g);
   if(c=='e')
   {
    POP(&g);
      }
   if(c=='t')
   {
    POP(&g);
   }
}
else cout<<"此语法错误"<<endl;

return 0;
}
void st()
{
    //SETNULL(&g);
if(EMPTY(&g))
{PUSH(&g,'$');
PUSH(&g,'E');
}
}
void main()
{
SETNULL(&g);
Input();
st();
for(int j=0;j<=l;j++)
   {
   cout<<g.data[j];
   }
   cout<<"                                     ";
   for(int i=save_j;i<=save_i;i++)
   {
    cout<<a[i];
   }
   cout<<endl;
while(!(g.data[g.top]=='$'&&a[save_j]=='$')&&!(a[save_j]=='i'&&a[save_j+1]=='i')&&!(a[save_j]=='+'&&a[save_j+1]=='*')&&!(a[save_j+1]=='+'&&a[save_j]=='*')&&!(g.data[g.top]==')'&&a[save_j]=='$'))
{
  //while(!(g.data[g.top]=='$'&&a[save_j]=='$'))
  {
  LL1();
  for(int j=0;j<=l;j++)
   {
   cout<<g.data[j];
   }
   cout<<"                                     ";
      for(int i=save_j;i<=save_i;++i)
   {
   cout<<a[i];
   }
   cout<<endl;
  }

   }
  //cout<<"此处错误不应该出现双运算符"<<endl;

}

#8

谢谢!真心的谢谢帮忙.

#1