C语言实现LL(1)文法分析器示例：表达式解析

/***********************************************************
*       实验三  LL(1)分析表演示程序模板(教学实验用)
*
*    采用LL(1)表分析法实现表达式文法的语法检验。
*    (0)E ->TX
*    (1)X ->+TX (2)X ->-TX (3)X ->ε
*    (4)T ->FY
*    (5)Y ->*FY (6)Y ->/FY (7)Y ->ε 
*  (8)F ->(E) (9)F ->i
*  思路：其中i指代数字。先通过词法分析，将实际数字识别
*          用i替代，再进行分析。如
*          3.14*2 =>  i*i
*
*                                 果冻不熟 2012.6.23
*                                 Q群：68229430
************************************************************/

#include <string.h>
#include <stdio.h>
#include <stdlib.h>

#define MAXPROD 10 //产生式候选数目

typedef struct production{
char LeftCode; //产生式左部
char rightCodes[5];    //产生式右部
char allCodes[10];         //产生式
}production;

char stack[200];     //堆栈体

char inputstr[100];             //输入串          
production prods[MAXPROD]; //产生式集

void init ()
{
prods[0].LeftCode = 'E';  strcpy(prods[0].rightCodes,"TX\0");  strcpy(prods[0].allCodes,"E ->TX\0");
prods[1].LeftCode = 'X';  strcpy(prods[1].rightCodes,"+TX\0"); strcpy(prods[1].allCodes,"X ->+TX\0");
prods[2].LeftCode = 'X';  strcpy(prods[2].rightCodes,"-TX\0"); strcpy(prods[2].allCodes,"X ->-TX\0");
prods[3].LeftCode = 'X';  strcpy(prods[3].rightCodes,"ε\0");  strcpy(prods[3].allCodes,"X ->ε");
prods[4].LeftCode = 'T';  strcpy(prods[4].rightCodes,"FY\0");  strcpy(prods[4].allCodes,"T ->FY\0");
prods[5].LeftCode = 'Y';  strcpy(prods[5].rightCodes,"*FY\0"); strcpy(prods[5].allCodes,"Y ->*FY\0");
prods[6].LeftCode = 'Y';  strcpy(prods[6].rightCodes,"/FY\0"); strcpy(prods[6].allCodes,"Y ->/FY\0");
prods[7].LeftCode = 'Y';  strcpy(prods[7].rightCodes,"ε\0");  strcpy(prods[7].allCodes,"Y ->ε\0");
prods[8].LeftCode = 'F';  strcpy(prods[8].rightCodes,"(E)\0"); strcpy(prods[8].allCodes,"F ->(E)\0");
prods[9].LeftCode = 'F';  strcpy(prods[9].rightCodes,"i\0");   strcpy(prods[9].allCodes,"F ->i\0");
}

//堆栈操作-->产生式倒着存入进栈
int StackPush (int *top, production prod)
{
int len, i;

if ((*top) < 0)
{
printf("\n\n\t\t\t不知道的错误！\n\n");
exit(0);
}


len = strlen(prod.rightCodes);