printf() - 打印错误的顺序.单独打印时工作

Question

我正在尝试以常见的日志格式打印一些文本.

printf("%s - - [%s] %s %d %zu\n", ip, _time, row, statuscode, size);

问题是订单混乱了.输出是:

200 1511 - - [20/Sep/2017:13:07:32 +0200] GET / HTTP/1.1

我认为(1511)是ip印刷品.不知道为什么.

当我像这样打印它们时:

printf("1. %s\n", ip);
printf("2. %s\n", _time);
printf("3. %s\n", row);
printf("4. %d\n", statuscode);
printf("5. %zu\n", size);

它像预期的那样工作:

1. 127.0.0.1
2. 20/Sep/2017:13:11:24 +0200
3. GET / HTTP/1.1
4. 200
5. 151

当我statuscode因某些原因添加时,问题似乎开始了.我不知道为什么.任何帮助表示赞赏.

这是我使用prinft()的函数:

static void handlelogging(char* method, struct sockaddr_storage client_addr, size_t size, char* row, int statuscode) {

char* ip;
char _time[80];
struct tm *info;
time_t rawtime;

time(&rawtime);
info = localtime(&rawtime);
strftime(_time, 80,"%d/%b/%Y:%H:%M:%S %z", info)
ip = getip(client_addr);

//Print goes here, see above.
}

正如有人说这可能是ip变量的问题:

static char* getip(struct sockaddr_storage client_addr) {
    char ipstr[20];
    struct sockaddr_in *s;


    s = (struct sockaddr_in *) &client_addr;

    /**
     * Converts network address (s) in the IPV_4 family into a string.
     */
    return strdup(inet_ntop(AF_INET, &s->sin_addr, ipstr, sizeof ipstr));
 }

Answer 1

我认为问题来自row哪个包含一个\r角色.

只需修改它 strchr(...)

/* replace all '\r' by 'R' in row */
char *p = strchr(row, '\r');
while(p)
{ 
    *p = 'R';
    p = strchr(row, '\r');
}