linux socket recv返回,Linux socket：如何使send()等待recv()

我正在使用TCP协议创建一个简单的客户端 – 服务器应用程序.

我知道默认情况下. recv()将阻塞,直到另一方调用send()到此套接字.

但是有可能send()阻塞自己,直到另一方已经recv()编辑了msg而不是将send()保持到传出队列,然后找到另一侧recv()得到了一大堆由多个发送的消息发送()■

换一种说法.是否有可能让每个send()等待另一方的recv()才能调用另一个send()？

说清楚我的问题.我将在这里发布一个简单的代码：

client.c

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

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

{

int sockfd = 0;

char sendBuff[1024];

struct sockaddr_in serv_addr;

int i;

if(argc != 2)

{

printf("\n Usage: %s \n",argv[0]);

return 1;

}

if((sockfd = socket(AF_INET,SOCK_STREAM,0)) < 0)

{

printf("\n Error : Could not create socket \n");

return 1;

}

memset(&serv_addr,'0',sizeof(serv_addr));

serv_addr.sin_family = AF_INET;

serv_addr.sin_port = htons(5000);

if(inet_pton(AF_INET,argv[1],&serv_addr.sin_addr)<=0)

{

printf("\n inet_pton error occured\n");

return 1;

}

if( connect(sockfd,(struct sockaddr *)&serv_addr,sizeof(serv_addr)) < 0)

{

printf("\n Error : Connect Failed \n");

return 1;

}

do{

memset(sendBuff,'\0',sizeof(sendBuff));

sprintf(sendBuff,"This is line %d",i);

send(sockfd,sendBuff,strlen(sendBuff),0);

//sleep(1);

}while(++i<100);

return 0;

}

server.c

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

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

{

int listenfd = 0,connfd = 0;

struct sockaddr_in serv_addr;

char sendBuff[1025];

char recvBuff[100];

int i = 0;

listenfd = socket(AF_INET,0);

memset(&serv_addr,sizeof(serv_addr));

memset(sendBuff,sizeof(sendBuff));

serv_addr.sin_family = AF_INET;

serv_addr.sin_addr.s_addr = htonl(INADDR_ANY);

serv_addr.sin_port = htons(5000);

bind(listenfd,(struct sockaddr*)&serv_addr,sizeof(serv_addr));

listen(listenfd,10);

connfd = accept(listenfd,(struct sockaddr*)NULL,NULL);

do{

memset(recvBuff,sizeof(recvBuff));

recv(connfd,recvBuff,sizeof(recvBuff),0);

printf( "%s\n",recvBuff);

}while(++i<100);

return 0;

}

我期望在服务器端的结果是打印：

This is line 0

This is line 1

This is line 2

This is line 3

...

但是,实际结果是这样的：

This is line 0

This is line 1This is line 2This is line3This is line 4

This is line 5This is line 6This is line 7This is line 8This is line 9This is line 10

This is line 11This is line 12...

但是这很容易解释：当客户端发出send()时,它没有等待服务器端的recv()完成,并且由于某种原因,服务器端recv()循环比客户端发送慢().因此,客户端上的几个send()可以堆叠在一起并由服务器作为整体接收. (我的解释是对的吗？)

实际上似乎有一个非常愚蠢和松散的解决方案.只需在循环中的每个send()之后添加一个sleep(1)(就像我注释掉的那样).我知道这将使代码效率非常低,并且如果recv()循环将花费更长的时间来实现一些其他复杂的操作(当程序变大时这显然是不可预测的),这将花费超过1秒.此解决方案失败！

那么是否有更好的方法让双方相互通信以确保由单个recv()接收的单个send()发送的消息？