| 网站首页 | 业界新闻 | 小组 | 威客 | 人才 | 下载频道 | 博客 | 代码贴 | 在线编程 | 编程论坛
欢迎加入我们,一同切磋技术
用户名:   
 
密 码:  
共有 429 人关注过本帖
标题:C语言写的一个parser 编译没问题 运行不能正常parse 求助!!!
取消只看楼主 加入收藏
alberm
Rank: 1
等 级:新手上路
帖 子:1
专家分:0
注 册:2012-10-24
收藏
 问题点数:0 回复次数:0 
C语言写的一个parser 编译没问题 运行不能正常parse 求助!!!
//  Grammar
//  SID a simple C identifier of unlimited length
//  STRLIT is a simple C-type literal string string
//  IEXPR is an integer expression

//  strexpr: SID ( term ) | str
//  term:    str , term | str | str + str
//  str:     SID str1 | SID() str1 | STRLIT str1
//  str1:    [ content ] str1 | epsilon
//  content: IEXPR | IEXPR : IEXPR | IEXPR : | : IEXPR

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

#define isliteral(x) ((x)=='\"')
#define isdigit(x) ((x)>='0'&&(x)<='9')
#define isfirst(x) (((x)>='A'&&(x)<='Z')||((x)>='a'&&(x)<='z')||((x)=='_'))
#define isnext(x)  (isfirst(x)||((x)>='0'&&(x)<='9'))
#define isws(x) ((x)==' '||(x)=='\n'||(x)=='\t')
#define SavePosition() spp1 = spp
#define RestorePosition() spp = spp1
 

int spp;
unsigned long token_ival;

// define token names

#define EOI  0
#define SID  256
#define STRLIT  SID+1
#define IEXPR  STRLIT+1
#define ERR  IEXPR+1

//function Parse
int Parse(char* s){
int token,spp2;
spp=0;
if(!strexpr(s)){
    return 0;
}
spp2=spp;
token=NextToken(s);
if(token==ERR){
    printf("Error occurred at position %d\n",spp2);
    exit(1);
    }
if(token!=EOI){
    printf("Reduntant stuff at position %d\n",spp2);
    exit(1);
    }
return 1;
}//end Parse

//strexpr: SID ( term ) | str
int strexpr(char* s){
int spp1,spp2,token;
spp1=0;
SavePosition();
token=NextToken(s);
if(token==SID){
   token=NextToken(s);
   if (token=='('){
   if (!term(s)){
   RestorePosition();
   return 0;
   }
   spp2=spp;
   if(NextToken(s)!=')'){
       printf("Expected token ) at position %d\n",spp2);
       exit(1);
       }
   return 1;
    }
    RestorePosition();
    return 0;
   }
   RestorePosition();
   if(!str(s)){
       RestorePosition();
       return 0;
       }
   return 1;
}//end strexpr

//  term:    str , term | str | str + str
int term(char* s){
int spp1,spp2,token;
spp1=0;
SavePosition();
if(!str(s)){
    RestorePosition();
    return 0;
    }
spp2=spp;
token=NextToken(s);
if(token==','){
    if (!term(s)){
        RestorePosition();
        return 0;
        }
    return 1;
    }
if(token=='+'){
    if (!str(s)){
        RestorePosition();
        return 0;
        }
    return 1;
    }
spp=spp2;
return 1;
}//end term

//  str:     SID str1 | SID() str1 | STRLIT str1
int str(char* s){
int spp1,spp2,token;
spp2=0;
SavePosition();
token=NextToken(s);
if(token==SID){
    spp2=spp;
    token=NextToken(s);
if(token=='('){
    token=NextToken(s);
    if(token==')'){
        if(!str1(s)){
            RestorePosition();
            return 0;
            }
        return 1;
        }
     spp=spp2;
     return 0;
    }
    spp=spp2;
    if(!str1(s)){
       RestorePosition();
       return 0;
       }
    return 1;
    }
//RestorePosition();
//token=NextToken(s);
if(token==STRLIT){
    if(!str1(s)){
        RestorePosition();
        return 0;
        }
    return 1;
    }
 //RestorePosition();
 return 0;
}//end str

//  str1:    [ content ] str1 | epsilon
int str1(char* s){
int spp1,token;
SavePosition();
token=NextToken(s);
if(token=='['){
    if(!content(s)){
           RestorePosition();
        return 1;
        }
    token=NextToken(s);
    if(token==']'){
        if(!str(s)){
        RestorePosition();
        return 1;
        }
        return 1;
      }
      //RestorePosition();
      return 1;
    }
//RestorePosition();
return 1;
}//end str1

//  content: IEXPR | IEXPR : IEXPR | IEXPR : | : IEXPR
int content(char* s){
int spp1,spp2,spp3,spp4,token;
SavePosition();
token=NextToken(s);
if(token==IEXPR){
   spp2=spp;
   token=NextToken(s);
   if(token==':'){
      spp3=spp;
      token=NextToken(s);
      if(token==IEXPR){
         return 1;
         }
      spp=spp3;
      return 1;
    }
    spp=spp2;
    return 1;
}
//RestorePosition();
//token=NextToken(s);
if(token==':'){
   spp4=spp;
   token=NextToken(s);
   if(token==IEXPR){
      return 1;
      }
      printf("Expected token IEXPR at position %d\n",spp4);
      exit(1);
    }
if(token==EOI){
    RestorePosition();
    return 0;
    }
    printf("Expected token IEXPR or : at position %d\n",spp1);
    exit(1);
}//end content

//function NextToken
int NextToken(char* s){
    while(isws(s[spp])){
        spp++; }                 //eat ws
if((s[spp])=='\0'){
    spp++;
    return EOI;
}else if(s[spp]==','){
    spp++;
    return ',';
}else if(s[spp]=='['){
    spp++;
    return '[';
}else if(s[spp]==']'){
    spp++;
    return ']';
}else if(s[spp]==':'){
    spp++;
    return ':';
}else if(s[spp]=='('){
    spp++;
    return '(';
}else if(s[spp]==')'){
    spp++;
    return ')';
}else if(s[spp]=='+'){
    spp++;
    return '+';
}else if(isfirst(s[spp])){
    spp++;                         //eat first char
    while(isnext(s[spp])){
          spp++;                   //eat all the following char
          return SID;
          }
}else if(isliteral(s[spp])){
          spp++;                      //eat "
          while(isnext(s[spp])){
          spp++;                   //eat all the following char except " in the end
      if(isliteral(s[spp])){      
          spp++;                   //eat "   
          return STRLIT;
        }
    }        
}else if(isdigit(s[spp])){
          token_ival=(s[spp]-'0');
          spp++;                   //eat first integer
          while(isdigit(s[spp])){
          token_ival=token_ival*10+(s[spp]-'0');
          spp++;                   //eat all the following integers
          }
          return IEXPR;
}
return ERR;
}//end NextToken

int main()
{
  char string[]="abc";
  if (Parse(string)) {
    printf("OK\n");
  }else
    printf("SYNTAX ERROR\n");
  return 0;
}
搜索更多相关主题的帖子: include content expression 
2012-10-24 08:05
快速回复:C语言写的一个parser 编译没问题 运行不能正常parse 求助!!!
数据加载中...
 
   



关于我们 | 广告合作 | 编程中国 | 清除Cookies | TOP | 手机版

编程中国 版权所有,并保留所有权利。
Powered by Discuz, Processed in 0.013230 second(s), 8 queries.
Copyright©2004-2024, BCCN.NET, All Rights Reserved