为什么 yacc 程序会出现分段错误？

我正在尝试从此HOW-TO运行示例（4.3。处理配置文件） 。这是作为 L1 语言输入的代码：

zone "." {
        type hint;
        file "/etc/bind/db.root";
};

这些是我正在使用的来源：

LEX-файл (example6.l)
%{
#include <stdio.h>
#include "y.tab.h"
%}

%%

zone            return ZONETOK;
file            return FILETOK;
[a-zA-Z][a-zA-Z0-9]*    yylval=strdup(yytext); return WORD;
[a-zA-Z0-9\/.-]+        yylval=strdup(yytext); return FILENAME;
\"                      return QUOTE;
\{                      return OBRACE;
\}                      return EBRACE;
;                       return SEMICOLON;
\n                      /* ignore EOL */;
[ \t]+                  /* ignore whitespace */;
%%

下面是一个带有语法 (yacc) 的文件 (example6.y)

%{
#include <stdio.h>
#include <string.h>

#define YYSTYPE char *

int yydebug=0;//пока без дебага

void yyerror(const char *str)
{
    fprintf(stderr,"error: %s\n",str);
}

int yywrap()
{
    return 1;
}

main()
{
    yyparse();
}

%}

%token WORD FILENAME QUOTE OBRACE EBRACE SEMICOLON ZONETOK FILETOK

%%

commands:
    |    
    commands command SEMICOLON
    ;


command:
    zone_set 
    ;

zone_set:
    ZONETOK quotedname zonecontent
    {
        printf("Complete zone for '%s' found\n",$2);
    }
    ;

zonecontent:
    OBRACE zonestatements EBRACE 

quotedname:
    QUOTE FILENAME QUOTE
    {
        $$=$2;
    }
    ;

zonestatements:
    |
    zonestatements zonestatement SEMICOLON
    ;

zonestatement:
    statements
    |
    FILETOK quotedname 
    {
        printf("A zonefile name '%s' was encountered\n", $2);
    }
    ;

block: 
    OBRACE zonestatements EBRACE SEMICOLON
    ;

statements:
    | statements statement
    ;

statement: WORD | block | quotedname

我正在编译上述文件：

yacc -d example6.y
lex (example6.l)
cc lex.yy.c y.tab.c -o example6

我发现了分段错误

elvin@ubuntu:~/Documents/myfirstlex/conf_file$ ./example6
zone "." {
        type hint;
        file "/etc/bind/db.root";
};Segmentation fault (core dumped)
elvin@ubuntu:~/Documents/myfirstlex/conf_file$

试图找出错误，我在调试模式下运行

elvin@ubuntu:~/Documents/myfirstlex/conf_file$ ./example6
Starting parse
Entering state 0
Reducing stack by rule 1 (line 30):
-> $$ = nterm commands ()
Stack now 0
Entering state 1
Reading a token: zone "." {
        type hint;
        file "/etc/bind/db.root";
        type hint;
};Next token is token ZONETOK ()
Shifting token ZONETOK ()
.............................
Stack now 0 1 3 7 10 13 15
Entering state 18
Reducing stack by rule 10 (line 65):
$1 = token FILETOK ()
$2 = nterm quotedname ()
Segmentation fault (core dumped)

你能解释一下可能是什么问题吗？如果 strdup 是问题所在，如何避免使用它？