服务器端的select()具有多个客户端

时间:2013-03-23 18:51:04

标签: c client-server

我正在研究客户端服务器代码,我在客户端运行两个线程(批量和零售),我在服务器上有两个端口来接收数据。我想在服务器端使用select()但由于在“sd1 - 哪个是批量端口”从未收到哪些数据而做了一些错误。如果有人可以帮我解决当前代码的问题,我将非常感激。谢谢!

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

fd_set readfds, activefds;

// SOCKET CREATION RETAIL
sd=socket(AF_INET, SOCK_DGRAM, 0);
if(sd<0) {
printf("error creating retail socket \n");
return -1;
}   

// SOCKET CREATION BULK
sd1=socket(AF_INET, SOCK_DGRAM, 0);
if(sd1<0) {
printf("error creating retail socket \n");
return -1;
}   

// BIND RETAIL PORT 
servAddr.sin_family = AF_INET;
servAddr.sin_addr.s_addr = htonl(INADDR_ANY);
servAddr.sin_port = htons(RETAIL_PORT);
rc = bind (sd, (struct sockaddr *) &servAddr,sizeof(servAddr));
if(rc<0) {
printf("error binding retail port number %d \n",RETAIL_PORT);
return -1;
}

// BIND BULK PORT
servAddr.sin_family = AF_INET;
servAddr.sin_addr.s_addr = htonl(INADDR_ANY);
servAddr.sin_port = htons(BULK_PORT);
rc1 = bind (sd1, (struct sockaddr *) &servAddr,sizeof(servAddr));
if(rc1<0) {
printf("error binding bulk port number %d \n",BULK_PORT);
return -1;
}

int nfds;
if(sd>sd1){
 nfds = sd;}
else{
 nfds = sd1;}

FD_ZERO(&activefds);
FD_SET(sd, &activefds);
FD_SET(sd1, &activefds);    

  while(1){

   readfds = activefds;

   if (select(nfds+1, &readfds, NULL, NULL, NULL) < 0){
     perror("select");
     break;
    }

       for(i=0;i<nfds;i++){

   if(FD_ISSET(i, &readfds)){
   printf("sd = %d\n\n\n",sd);  // sd = 3 getting printed here
    if(i == sd){
      n = recvfrom(sd, &requested_retail_cases, 4, 0, (struct sockaddr *) &cliAddr, &cliLen);
           printf("in retail\n");
          }

  else if(FD_ISSET(i, &readfds)){

    if(i == sd1){

    n1 = recvfrom(sd1, &requested_bulk_cases, 4, 0, (struct sockaddr *) &cliAddr, &cliLen);
  }

    }
 }
}
 return 1;
 }

1 个答案:

答案 0 :(得分:0)

更改

if(FD_ISSET(i, &readfds)){
   if(i == sd){
      ...
   }
else if(FD_ISSET(i, &readfds)){
   if(i == sd1){
      ...
   }
}

if(FD_ISSET(i, &readfds)){
   if(i == sd){
      ...
   }
   else if(i == sd1){
      ...
   }
}

在代码中,第二个外部if永远不会到达。你实际上有if (x) then y; else if (x) then z;显然,第二个if永远不会到达。