看了action的讲解， 没搞清楚，求助。

下面的程序， 当输入[[]]时， 为什么输出on_open_array一次，而输出on_close_array两次。

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

%%{
    machine openclose;
    write data;
}%%

void test(char *str)
{
    int cs;
    char *p = str, *pe = str + strlen(str);
    bool found_open_obj = false;
    bool found_open_array = false;
    bool found_close_obj = false;
    bool found_close_array = false;
    
    %%{
        action on_open_obj {
            printf("on_open_obj\n");
            found_open_obj = true;
        }  
        action on_open_array {
            printf("on_open_array\n");
            found_open_array = true;
        }  

        action on_close_obj {
            printf("on_close_obj\n");
            found_close_obj = true;
        }  
        action on_close_array {
            printf("on_close_array\n");
            found_close_array = true;
        }  

        main := (space*) ( '{'@on_open_obj | '['@on_open_array )  any* ( '}'@on_close_obj | ']'@on_close_array ) space* '\n';

        # Initialize and execute.
        write init;
        write exec;
    }%%

    if ( cs < openclose_first_final )
    {
        fprintf( stderr, "there was an syntax error\n" );
    }
    else
    {
        if(found_open_obj && found_close_obj)
        {
            fprintf( stderr, "startwith open obj, endwith close obj\n" );
        }
        else if(found_open_array && found_close_array)
        {
            fprintf( stderr, "startwith open array, endwith close array\n" );
        }
        else
        {
            fprintf( stderr, "wrong format\n" );
        }
    }
};

int main(int argc, char *argv[])
{
    char buf[1024];

    while (fgets(buf, sizeof(buf), stdin) != 0) {
        test(buf);
    }
    return 0;
}
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