我使用 Windows 10 和 C (visual studios 2015) 编写了套接字服务器代码,但我认为绑定(bind)或监听代码是错误的。
当我执行我的代码时,它不会等到客户端连接。打印出“winsock初始化成功”和“创建socket成功”后结束。
请帮帮我
#include <stdio.h>
#include <winsock2.h>
int main(int argc, char *argv[]) {
WSADATA wsaData;
struct sockaddr_in address_of_server;
struct sockaddr_in address_of_client;
int socket_of_client;
int size_of_address_of_client = sizeof(address_of_client);
if (WSAStartup(MAKEWORD(2, 2), &wsaData) == 0) {
printf("winsock initialization success\n");
}
else {
printf("winsock initialization failure\n");
}
SOCKET socket_of_server = socket(AF_INET, SOCK_STREAM, 0);
if (socket_of_server == -1) {
printf("creating socket failure\n");
}
else {
printf("creating socket success\n");
}
memset(&address_of_server, 0, sizeof(address_of_server));
address_of_server.sin_family = AF_INET;
address_of_server.sin_addr.s_addr = htonl(INADDR_ANY);
address_of_server.sin_port = htons(atoi(10000));
bind(socket_of_server, (struct sockaddr*)&address_of_server, sizeof(address_of_server));
listen(socket_of_server, 5);
socket_of_client = accept(socket_of_server, (struct sockaddr*)&address_of_client, &size_of_address_of_client);
WSACleanup();
}
最佳答案
在 listen() 之后你需要调用 accept() 来获得一个新的连接套接字。 listen() 只是开始监听,并不等待任何客户端连接。
关于c - 使用 Windows 10 和 C (Visual Studio 2015) 进行套接字编程,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50607736/