这不是一个真正的答案,但我制作了一个小服务器脚本来kqueue
解释问题:
#include <stdio.h> // fprintf
#include <sys/event.h> // kqueue
#include <netdb.h> // addrinfo
#include <arpa/inet.h> // AF_INET
#include <sys/socket.h> // socket
#include <assert.h> // assert
#include <string.h> // bzero
#include <stdbool.h> // bool
#include <unistd.h> // close
int main(int argc, const char * argv[])
{
/* Initialize server socket */
struct addrinfo hints, *res;
int sockfd;
bzero(&hints, sizeof(hints));
hints.ai_family = AF_INET;
hints.ai_socktype = SOCK_STREAM;
assert(getaddrinfo("localhost", "9090", &hints, &res) == 0);
sockfd = socket(AF_INET, SOCK_STREAM, res->ai_protocol);
assert(sockfd > 0);
{
unsigned opt = 1;
assert(setsockopt(sockfd, SOL_SOCKET, SO_REUSEADDR, &opt, sizeof(opt)) == 0);
#ifdef SO_REUSEPORT
assert(setsockopt(sockfd, SOL_SOCKET, SO_REUSEPORT, &opt, sizeof(opt)) == 0);
#endif
}
assert(bind(sockfd, res->ai_addr, res->ai_addrlen) == 0);
freeaddrinfo(res);
/* Start to listen */
(void)listen(sockfd, 5);
{
/* kevent set */
struct kevent kevSet;
/* events */
struct kevent events[20];
/* nevents */
unsigned nevents;
/* kq */
int kq;
/* buffer */
char buf[20];
/* length */
ssize_t readlen;
kevSet.data = 5; // backlog is set to 5
kevSet.fflags = 0;
kevSet.filter = EVFILT_READ;
kevSet.flags = EV_ADD;
kevSet.ident = sockfd;
kevSet.udata = NULL;
assert((kq = kqueue()) > 0);
/* Update kqueue */
assert(kevent(kq, &kevSet, 1, NULL, 0, NULL) == 0);
/* Enter loop */
while (true) {
/* Wait for events to happen */
nevents = kevent(kq, NULL, 0, events, 20, NULL);
assert(nevents >= 0);
fprintf(stderr, "Got %u events to handle...\n", nevents);
for (unsigned i = 0; i < nevents; ++i) {
struct kevent event = events[i];
int clientfd = (int)event.ident;
/* Handle disconnect */
if (event.flags & EV_EOF) {
/* Simply close socket */
close(clientfd);
fprintf(stderr, "A client has left the server...\n");
} else if (clientfd == sockfd) {
int nclientfd = accept(sockfd, NULL, NULL);
assert(nclientfd > 0);
/* Add to event list */
kevSet.data = 0;
kevSet.fflags = 0;
kevSet.filter = EVFILT_READ;
kevSet.flags = EV_ADD;
kevSet.ident = nclientfd;
kevSet.udata = NULL;
assert(kevent(kq, &kevSet, 1, NULL, 0, NULL) == 0);
fprintf(stderr, "A new client connected to the server...\n");
(void)write(nclientfd, "Welcome to this server!\n", 24);
} else if (event.flags & EVFILT_READ) {
/* sleep for "processing" time */
readlen = read(clientfd, buf, sizeof(buf));
buf[readlen - 1] = 0;
fprintf(stderr, "bytes %zu are available to read... %s \n", (size_t)event.data, buf);
sleep(4);
} else {
fprintf(stderr, "unknown event: %8.8X\n", event.flags);
}
}
}
}
return 0;
}
每次客户端发送内容时,服务器都会经历 4 秒的“滞后”。(我夸大了一点,但对于测试来说相当合理)。那么如何解决这个问题呢?我认为具有自己的工作线程(池)kqueue
作为可能的解决方案,然后不会发生连接延迟。(每个工作线程读取某个“范围”的文件描述符)