TCP server有关缓冲区的问题

[qq_38790716]

//server.cpp
出问题的一段：

printf("==========waiting for client's request==========\n");
    if ((acceptfd = accept(listenfd, (struct sockaddr*)NULL, NULL)) == -1) {
        printf("accept socket error: %s(errno: %d)\n", strerror(errno), errno);
    }
    else
    {
        printf("accept sucess\n");
        int rcvbuf_len = 0;
        socklen_t len = sizeof(rcvbuf_len);
        if (getsockopt(acceptfd, SOL_SOCKET, SO_RCVBUF, &rcvbuf_len, &len) < 0) {
            perror("getsockopt: ");
        }
        printf("the recv buf len: %d\n", rcvbuf_len);
    }
    char recvmsg[246988] = {0};
    ssize_t totallen = 0;
    while (1) {
        sleep(1);
        ssize_t readlen = read(acceptfd, recvmsg, sizeof(recvmsg));
        printf("readLen: %d\n", readlen);
        if (readlen < 0) {
            perror("read: ");
            return -1;
        }
        else if (readlen == 0) {
            printf("read finish: len = % ld\n", totallen);
            close(acceptfd);
            return 0;
        }
        else
        {
            totallen += readlen;
        }
    }

//client.cpp
客户端向服务端发送4个包，每个包中发送246988Byte：

ssize_t writelen;
    char sendmsg[246988] = {0};
    int count = 0;
    while (1)
    {
        ++count;
        if (count == 5) {
            return 0;
        }
        writelen = write(connfd, sendmsg, sizeof(sendmsg));
        if (writelen < 0) {
            printf("write error\n");
            close(connfd);
            return 0;
        }
        else
        {
            printf("write success, writelen: %d\n", writelen);
        }
    }

然后所得结果：




然后我测了一下ubuntu 18.04的默认缓冲区是87380Byte，然后就想知道这个为什么是1062000？

[hou09tian]

MSDN关于getsockopt()函数的解释中，对SO_RCVBUF有这么一段描述。
The total per-socket buffer space reserved for receives. This is unrelated to SO_MAX_MSG_SIZE and does not necessarily correspond to the size of the TCP receive window.
可以看到，将getsockopt()函数的参数设置为SO_RCVBUF之后，获取的套接字缓冲区与TCP接收窗口无关。

[qq_38790716]

但是按理说这个应该得到默认缓冲区的大小，，，我看书上得到的也是这个结果，所以就怀疑是不是有点问题。。。