我已经为rsyslog配置了以下过滤器,将一些SSH消息定向到本地系统上的特定TCP端口5000,以便5000上运行的服务将进一步处理SSH消息.
if $fromhost-ip == '127.0.0.1' and ( ($msg contains 'SSH') and ($msg contains 'Test') ) then @@127.0.0.1:5000
一切似乎都很好,但消息没有重定向到端口5000,如果我们将消息定向到UDP端口,它工作正常.
以下是指向UDP端口的消息的过滤器.
if $fromhost-ip == '127.0.0.1' and ( ($msg contains 'SSH') and ($msg contains 'Test') ) then @127.0.0.1:5000
能不能让我知道,为什么TCP端口不起作用,UDP端口工作.
我写了一个小代码，当无法连接到 postgres 数据库时，使用 C Api 将消息发送到系统日志。
int main ( int argc, char **argv )
{
PGconn *psql;
PGresult *res;
int flag = 0;
openlog ( "postgres", LOG_NDELAY, LOG_SYSLOG );
psql = PQconnectdb("hostaddr = '127.0.0.0' port = '5432' dbname = 'RtpDb' user = 'rtp_user_99' password = 'rtp_user' connect_timeout = '10'");
if ( PQstatus(psql) != CONNECTION_OK )
{
//Send an event to syslog for DB Connection Failure
syslog (LOG_EMERG, "%s", PQerrorMessage(psql) )
}
closelog ();
PQclear(res);
PQfinish(psql);
}
当 postgres 数据库连接失败时，即使在 openlog …
请告诉我堆栈和堆之间的区别与下面的代码
int main()
{
int arr[3];
int *a;
arr [5] = 6; // out of bound but it will not give error.
arr [3000] = 8 ; //SIGSEGV
a = malloc (sizeof (int));
a[4] = 6;
a[4000] = 8; //No error
}
我知道arr是一个静态数组,当我执行arr [3000]时会访问其他进程的地址,这会给出SIGSEGV错误.但我不明白为什么[4000]不会给我任何运行时错误,即SIGSEGV信号.
谢谢
以下是代码段.
#define TABLE_DELIMITER "::"
int parse_n_store ( char *line )
{
int i = 0;
char *p = NULL;
CPTR sensor_number = NULL , event_catagory = NULL, sensor_type = NULL, event_state= NULL, assertion = NULL, message_number = NULL, short_text = NULL;
for (p = strtok(line,TABLE_DELIMITER); p != NULL; p = strtok(NULL, TABLE_DELIMITER), i++ )
{
if ( i == 0 )
sensor_number=p;
else if ( i == 1 )
sensor_type = p;
else if ( i == 2 )
event_catagory …下面是源代码的一小部分fclose导致错误的地方？此函数并不总是被调用，在某些特定条件下会调用此函数。
int write_into_file (char * file_name)
{
FILE * fp = NULL ;
if (file_name == NULL)
{
return FAIL ;
}
if ((fp = fopen (file_name , "r")) == NULL)
{
if ((fp = fopen (file_name, "w")) == NULL)
{
return FAIL ;
}
}
fclose (fp) ;
fp = NULL ;
return SUCESS;
}
我们将大小为 1024 的字符缓冲区传递给 file_name。请谁能告诉我为什么 fclose 会导致分段错误？
我是Perl剧本的新手.
我想对文件进行读写操作.我将以读写模式(+ <)打开文件,并写入文件.现在,我想读取以前写过的文件.以下是我的代码:
#!/usr/bin/perl
`touch file.txt`; #Create a file as opening the file in +< mode
open (OUTFILE, "+<file.txt") or die "Can't open file : $!";
print OUTFILE "Hello, welcome to File handling operations in perl\n"; #write into the file
$line = <OUTFILE>; #read from the file
print "$line\n"; #display the read contents.
当我显示阅读内容时,它显示一个空白行.但是文件"file.txt"有数据
Hello, welcome to File handling operations in perl
为什么我无法阅读内容.我的代码是错误的还是我错过了什么.
Below is my code snippet.
int main ( )
{
some instructions;
while ( 1 )
{
/* Block 1 : Starts*/
if ( selection == 1 )
{
ret = pthread_create ( &tid, NULL, &select_n_process_req, NULL );
if ( ret != 0 )
{
printf ("Error Creating Thread");
}
}
/* Block 1 : Ends*/
/* Block 2 : Starts*/
printf ("Sleeping for [%d]", retry_time * 60 * 60);
for ( i = 0; i < retry_time * 60 …为什么以下代码会出现分段错误？
int main()
{
char *t = "Working on RedHat Linux";
char *s;
s = malloc (8000 * sizeof(char));
memcpy(s,t,7000);
printf("s = %s\nt = %s\n",s,t);
free(s);
}
我为's'分配了8000bytes.并且仅将't'复制到s直到7000字节.虽然我为's'分配了8000个字节,为什么它会给出分段错误？