编译原理JAVA模拟LL(1)文法分析过程

{"T’","","ε/P","T’F/N","","ε/P","ε/P"},
{"F","ε/N","","",")E/N","",""},
{")","","","","","ε/N",""},
{"#","","","","","","acc"}
index1 = stack.length()-2;//
else
index1=stack.length()-1;
topStack=String.valueOf(stack.substring(index1,stack.length()));//栈顶元素
while(kong>=0)//去掉输入串表达式中的空格
{
inString.delete(kong,kong+1);
kong=inString.indexOf(" ");
}
StringBuffer stack=new StringBuffer("#E");//分析栈，初始放入E
inString.delete(0,1);//相当于读下一个元素
liutemp = new StringBuffer(first);
list.add(ll[x][0]+"->"+topIn+reverse(liutemp));
if(stack.toString().endsWith("\'")||stack.toString().endsWith("’"))
System.out.println("不能完整匹配！");
}
}
else if(last.equals("/P"))
{
System.out.println(id+"\t"+stack+getBlank(21-stack.length()-inString.length())+inString+getBlank(11-operation.length())+operation);
index1 = stack.length()-2;//重新设置index1值
else
index1=stack.length()-1;
id++;
}else {
System.out.println("分析表构造出错！");
};//ll(1)分析矩阵
System.out.println("LL(1)分析过程如下：");
System.out.println("\n序号\t分析栈"+getBlank(20)+" 输入数据"+getBlank(20)+"动作");
StringBuffer liutemp = null;
System.out.println("请输入一个语句:");
Scanner in = new Scanner(System.in);
String ins=in.nextLine().trim();
if(ins.indexOf("#")<0)//假如最后完了输入#号也没事
stack.delete(0,1);
inString.delete(0,1);
System.out.println("匹配成功！");
}
else
{
System.out.println(id+"\t"+stack+getBlank(21-stack.length()-inString.length())+inString+getBlank(6)+"error");
int ix=-1;
int length=0;
if(buffer.indexOf("'")<0&&buffer.indexOf("’")<0)
buf.append(buffer.reverse());
else{
while(buffer.length()>0){
System.out.println("输入的符号不符合规定文法！");
System.exit(0);
}else if(x!=0&&y!=0&&operation.length()==0)
{
System.out.println(id+"\t"+stack+getBlank(21-stack.length()-inString.length())+inString+inString+getBlank(6)+"error");
if(topStack.equals(ll[i][0]))
{
x=i;
break;
}
for(int i=1;i<ll[0].length;i++)//计算对应分析表的列数
if(topIn.equals(ll[0][i])th()>0)
topIn=String.valueOf(inString.charAt(0));//剩余输入串的第一个元素
else
topIn="";
for(int i=1;i<ll.length;i++)//计算对应分析表的哪一行
{
int id=1;
int index1=1;//记录栈最后一个非终结符的位置
String topStack,topIn,operation;
StringBuffer inString=null;
ArrayList<String> list = new ArrayList<String>();
String last=operation.substring(operation.length()-2,operation.length());//是否换行部分
if(first.equals("ε"))//如果是空字符，有不要加入栈
first="";
if(operation.equals("acc"))
System.exit(0);
}
}else if(y==0){
System.out.println(id+"\t"+stack+getBlank(21-stack.length()-inString.length())+inString+getBlank(6)+"error");
import java.util.ArrayList;
//lsh
import java.util.Scanner;
/**
*题目：LL(1)分析过程模拟
* @author:刘sh 自己独立完成,林科大
* time:4月22日
* 功能：
* 1、适应对满足LL（1）某一文法的输入串分析
{
y=i;
break;
}
operation=ll[x][y];//动作命令
if(operation.length()>=3)
{
String first=operation.substring(0,operation.length()-2);//替换部分
stack.replace(index1,index1+topStack.length(),first);//把栈顶元素替换为分析表中值
if(first.equals(""))
list.add(ll[x][0]+"->ε");
else{
System.out.println("输入符号串不完整！");
System.exit(0);
}
}
System.out.println("\n该语句自顶向下构建语法树过程：");
for(int i = 0;i<list.size();i++)
* 2、适应输入串中插入空格
* 3、有简单的错误提示
*假定文法如下：
* E→TE’
* E’→+TE’︱ε
* T→FT’
* T’→*FT’︱ε
* F→(E)︱i
*/
public class LL1Analysis {
public static void main(String a[])
ins+="#";
else
ins=ins.substring(0,ins.indexOf("#")+1);//截取#号在内的前部