为什么在工作交错时TCP写入延迟会更糟？

Question

我一直在分析TCP延迟(特别是write从一个小消息的用户空间到内核空间),以便对a的延迟有一些直觉write(确认这可能是特定于上下文的).我注意到测试之间存在很大的不一致,这对我来说似乎很相似,而且我很想知道差异来自何处.我知道微基准测试可能会有问题,但我仍然觉得我缺少一些基本的理解(因为延迟差异是~10倍).

设置是我有一个C++ TCP服务器接受一个客户端连接(来自同一CPU上的另一个进程),并且在与客户端连接时write,对套接字进行20次系统调用,一次发送一个字节.服务器的完整代码将在本文末尾复制.这是每次write使用的输出boost/timer(增加~1 mic的噪音):

$ clang++ -std=c++11 -stdlib=libc++ tcpServerStove.cpp -O3; ./a.out
18 mics
3 mics
3 mics
4 mics
3 mics
3 mics
4 mics
3 mics
5 mics
3 mics
...

我可靠地发现第一个write明显慢于其他人.如果我write在计时器中包装10,000个呼叫,平均每个2微秒write,但第一个呼叫总是15个以上的话筒.为什么会出现这种"变暖"现象？

相关地,我运行了一个实验,在每个write调用之间我做一些阻止CPU工作(计算一个大的素数).这会导致所有的write调用是缓慢的:

$ clang++ -std=c++11 -stdlib=libc++ tcpServerStove.cpp -O3; ./a.out
20 mics
23 mics
23 mics
30 mics
23 mics
21 mics
21 mics
22 mics
22 mics
...

鉴于这些结果,我想知道在将字节从用户缓冲区复制到内核缓冲区的过程中是否存在某种批处理.如果多次write调用快速连续发生,它们是否会合并为一个内核中断？

特别是我正在寻找write将缓冲区从用户空间复制到内核空间需要多长时间的概念.如果有一些合并效应允许平均值write在我连续执行10,000次时只占用2个麦克风,那么得出write延迟是2个麦克风的结论将是不公平的乐观; 似乎我的直觉应该是每个write需要20微秒.对于write没有内核旁路的最低延迟(一个字节的原始调用),这看起来非常慢.

最后一条数据是,当我在计算机上的两个进程(TCP服务器和TCP客户端)之间设置乒乓测试时,我平均每次往返6个麦克风(包括a read,a write,以及移动)通过localhost网络).这似乎与上面看到的单个写入的20个麦克风延迟不一致.

TCP服务器的完整代码:

// Server side C/C++ program to demonstrate Socket programming
// #include <iostream>
#include <unistd.h>
#include <stdio.h>
#include <sys/socket.h>
#include <stdlib.h>
#include <netinet/in.h>
#include <netinet/tcp.h>
#include <string.h>
#include <boost/timer.hpp>
#include <unistd.h>

// Set up some blocking work.
bool isPrime(int n) {
    if (n < 2) {
        return false;
    }

    for (int i = 2; i < n; i++) {
        if (n % i == 0) {
            return false;
        }
    }

    return true;
}

// Compute the nth largest prime. Takes ~1 sec for n = 10,000
int getPrime(int n) {
    int numPrimes = 0;
    int i = 0;
    while (true) {
        if (isPrime(i)) {
            numPrimes++;
            if (numPrimes >= n) {
                return i;
            }
        }
        i++;
    }
}

int main(int argc, char const *argv[])
{
    int server_fd, new_socket, valread;
    struct sockaddr_in address;
    int opt = 1;
    int addrlen = sizeof(address);

    // Create socket for TCP server
    server_fd = socket(AF_INET, SOCK_STREAM, 0);

    // Prevent writes from being batched
    setsockopt(server_fd, SOL_SOCKET, TCP_NODELAY, &opt, sizeof(opt));
    setsockopt(server_fd, SOL_SOCKET, TCP_NOPUSH, &opt, sizeof(opt));
    setsockopt(server_fd, SOL_SOCKET, SO_SNDBUF, &opt, sizeof(opt));
    setsockopt(server_fd, SOL_SOCKET, SO_SNDLOWAT, &opt, sizeof(opt));

    address.sin_family = AF_INET;
    address.sin_addr.s_addr = INADDR_ANY;
    address.sin_port = htons(8080);

    bind(server_fd, (struct sockaddr *)&address, sizeof(address));

    listen(server_fd, 3);

    // Accept one client connection
    new_socket = accept(server_fd, (struct sockaddr *)&address, (socklen_t*)&addrlen);

    char sendBuffer[1] = {0};
    int primes[20] = {0};
    // Make 20 sequential writes to kernel buffer.
    for (int i = 0; i < 20; i++) {
        sendBuffer[0] = i;
        boost::timer t;
        write(new_socket, sendBuffer, 1);
        printf("%d mics\n", int(1e6 * t.elapsed()));

        // For some reason, doing some blocking work between the writes
        // The following work slows down the writes by a factor of 10.
        // primes[i] = getPrime(10000 + i);
    }

    // Print a prime to make sure the compiler doesn't optimize
    // away the computations.
    printf("prime: %d\n", primes[8]);

}

TCP客户端代码:

// Server side C/C++ program to demonstrate Socket programming
// #include <iostream>
#include <unistd.h>
#include <stdio.h>
#include <sys/socket.h>
#include <stdlib.h>
#include <netinet/in.h>
#include <netinet/tcp.h>
#include <string.h>
#include <unistd.h>

int main(int argc, char const *argv[])
{
    int sock, valread;
    struct sockaddr_in address;
    int opt = 1;
    int addrlen = sizeof(address);

    // We'll be passing uint32's back and forth
    unsigned char recv_buffer[1024] = {0};

    // Create socket for TCP server
    sock = socket(AF_INET, SOCK_STREAM, 0);

    setsockopt(sock, SOL_SOCKET, TCP_NODELAY, &opt, sizeof(opt));

    address.sin_family = AF_INET;
    address.sin_addr.s_addr = INADDR_ANY;
    address.sin_port = htons(8080);

    // Accept one client connection
    if (connect(sock, (struct sockaddr *)&address, (socklen_t)addrlen) != 0) {
        throw("connect failed");
    }

    read(sock, buffer_pointer, num_left);

    for (int i = 0; i < 10; i++) {
        printf("%d\n", recv_buffer[i]);
    }
}

我想有和没有标志TCP_NODELAY,TCP_NOPUSH,SO_SNDBUF和SO_SNDLOWAT,的想法,这可能会阻止配料(但我的理解是,内核缓冲区和网络之间出现这种配料,而不是用户缓冲区和内核缓冲区之间).

这是乒乓测试的服务器代码:

// Server side C/C++ program to demonstrate Socket programming
// #include <iostream>
#include <unistd.h>
#include <stdio.h>
#include <sys/socket.h>
#include <stdlib.h>
#include <netinet/in.h>
#include <netinet/tcp.h>
#include <string.h>
#include <boost/timer.hpp>
#include <unistd.h>

 __inline__ uint64_t rdtsc(void)
   {
uint32_t lo, hi;
__asm__ __volatile__ (
        "xorl %%eax,%%eax \n        cpuid"
        ::: "%rax", "%rbx", "%rcx", "%rdx");
__asm__ __volatile__ ("rdtsc" : "=a" (lo), "=d" (hi));
return (uint64_t)hi << 32 | lo;
 }

// Big Endian (network order)
unsigned int fromBytes(unsigned char b[4]) {
    return b[3] | b[2]<<8 | b[1]<<16 | b[0]<<24;
}

void toBytes(unsigned int x, unsigned char (&b)[4]) {
    b[3] = x;
    b[2] = x>>8;
    b[1] = x>>16;
    b[0] = x>>24;
}

int main(int argc, char const *argv[])
{
    int server_fd, new_socket, valread;
    struct sockaddr_in address;
    int opt = 1;
    int addrlen = sizeof(address);
    unsigned char recv_buffer[4] = {0};
    unsigned char send_buffer[4] = {0};

    // Create socket for TCP server
    server_fd = socket(AF_INET, SOCK_STREAM, 0);

    address.sin_family = AF_INET;
    address.sin_addr.s_addr = INADDR_ANY;
    address.sin_port = htons(8080);

    bind(server_fd, (struct sockaddr *)&address, sizeof(address));

    listen(server_fd, 3);

    // Accept one client connection
    new_socket = accept(server_fd, (struct sockaddr *)&address, (socklen_t*)&addrlen);
    printf("Connected with client!\n");

    int counter = 0;
    unsigned int x = 0;
    auto start = rdtsc();
    boost::timer t;

    int n = 10000;
    while (counter < n) {
        valread = read(new_socket, recv_buffer, 4);
        x = fromBytes(recv_buffer);
        toBytes(x+1, send_buffer);
        write(new_socket, send_buffer, 4);
        ++counter;
    }

    printf("%f clock cycles per round trip (rdtsc)\n",  (rdtsc() - start) / double(n));
    printf("%f mics per round trip (boost timer)\n", 1e6 * t.elapsed() / n);
}

这是乒乓测试的客户端代码:

// #include <iostream>
#include <unistd.h>
#include <stdio.h>
#include <sys/socket.h>
#include <stdlib.h>
#include <netinet/in.h>
#include <netinet/tcp.h>
#include <string.h>
#include <boost/timer.hpp>
#include <unistd.h>

// Big Endian (network order)
unsigned int fromBytes(unsigned char b[4]) {
    return b[3] | b[2]<<8 | b[1]<<16 | b[0]<<24;
}

void toBytes(unsigned int x, unsigned char (&b)[4]) {
    b[3] = x;
    b[2] = x>>8;
    b[1] = x>>16;
    b[0] = x>>24;
}

int main(int argc, char const *argv[])
{
    int sock, valread;
    struct sockaddr_in address;
    int opt = 1;
    int addrlen = sizeof(address);

    // We'll be passing uint32's back and forth
    unsigned char recv_buffer[4] = {0};
    unsigned char send_buffer[4] = {0};

    // Create socket for TCP server
    sock = socket(AF_INET, SOCK_STREAM, 0);

    // Set TCP_NODELAY so that writes won't be batched
    setsockopt(sock, SOL_SOCKET, TCP_NODELAY, &opt, sizeof(opt));

    address.sin_family = AF_INET;
    address.sin_addr.s_addr = INADDR_ANY;
    address.sin_port = htons(8080);

    // Accept one client connection
    if (connect(sock, (struct sockaddr *)&address, (socklen_t)addrlen) != 0) {
        throw("connect failed");
    }

    unsigned int lastReceived = 0;
    while (true) {
        toBytes(++lastReceived, send_buffer);
        write(sock, send_buffer, 4);
        valread = read(sock, recv_buffer, 4);
        lastReceived = fromBytes(recv_buffer);
    }
}

Answer 1

这里有一些问题。

为了更接近答案，您需要让客户端做两件事： 1. 接收所有数据。2. 记录每次读取的大小。我这样做是通过：

  int loc[N+1];
int nloc, curloc;
for (nloc = curloc = 0; curloc < N; nloc++) {
    int n = read(sock, recv_buffer + curloc, sizeof recv_buffer-curloc);
    if (n <= 0) {
            break;
    }
    curloc += n;
    loc[nloc] = curloc;
}
int last = 0;
for (int i = 0; i < nloc; i++) {
    printf("%*.*s ", loc[i] - last, loc[i] - last, recv_buffer + last);
    last = loc[i];
}
printf("\n");

并将 N 定义为 20（抱歉，教养），并更改服务器以一次写入一个字节。现在，当打印出如下内容时：

 a b c d e f g h i j k l m n o p q r s 

我们知道服务器正在发送 1 字节数据包；但是当它打印出类似的内容时：

 a bcde fghi jklm nop qrs 

我们怀疑服务器主要发送 4 字节数据包。

根本问题是 TCP_NODELAY 没有执行您怀疑的操作。Nagle 算法，当有未确认的发送数据包时累积输出；TCP_NODELAY 控制是否应用此功能。

不管TCP_NODELAY如何，你仍然是一个STREAM_SOCKET，这意味着N次写入可以合并为一次。插座正在为设备供电，但同时您也在为插座供电。一旦数据包 [ mbuf, skbuff, ...] 被提交到设备，套接字需要在下一个 write() 上创建一个新数据包。一旦设备准备好接收新数据包，套接字就可以提供它，但在此之前，该数据包将充当缓冲区。在缓冲模式下，写入速度非常快，因为所有必要的数据结构都可用[如评论和其他答案中提到的]。

您可以通过调整 SO_SNDBUF 和 SO_SNDLOWAT 套接字选项来控制此缓冲。但请注意，accept 返回的缓冲区不会继承所提供套接字的缓冲区大小。通过将 SNDBUF 减少到 1

输出如下：

abcdefghijklmnopqrst 
a bcdefgh ijkl mno pqrst 
a b cdefg hij klm nop qrst 
a b c d e f g h i j k l m n o p q r s t 

对应从默认开始，然后在后续连接时向服务器端依次添加：TCP_NODELAY、TCP_NOPUSH、SO_SNDBUF（=1）、SO_SNDLOWAT（=1）。每次迭代的时间增量都比前一次更平坦。

您的情况可能会有所不同，这是在 MacOS 10.12 上；我用 rdtsc() 将你的程序更改为 C++ 程序，因为我有信任问题。

/* srv.c */
// Server side C/C++ program to demonstrate Socket programming
// #include <iostream>
#include <unistd.h>
#include <stdio.h>
#include <sys/socket.h>
#include <stdbool.h>
#include <stdlib.h>
#include <netinet/in.h>
#include <netinet/tcp.h>
#include <string.h>
#include <unistd.h>

#ifndef N
#define N 20
#endif
int nap = 0;
int step = 0;
extern long rdtsc(void);

void xerror(char *f) {
    perror(f);
    exit(1);
}
#define Z(x)   if ((x) == -1) { xerror(#x); }

void sopt(int fd, int opt, int val) {
    Z(setsockopt(fd, SOL_SOCKET, opt, &val, sizeof(val)));
}
int gopt(int fd, int opt) {
    int val;
    socklen_t r = sizeof(val);
    Z(getsockopt(fd, SOL_SOCKET, opt, &val, &r));
    return val;
}

#define POPT(fd, x)  printf("%s %d ", #x, gopt(fd, x))
void popts(char *tag, int fd) {
    printf("%s: ", tag);
    POPT(fd, SO_SNDBUF);
    POPT(fd, SO_SNDLOWAT);
    POPT(fd, TCP_NODELAY);
    POPT(fd, TCP_NOPUSH);
    printf("\n");
}

void stepsock(int fd) {
     switch (step++) {
     case 7:
    step = 2;
     case 6:
         sopt(fd, SO_SNDLOWAT, 1);
     case 5:
         sopt(fd, SO_SNDBUF, 1);
     case 4:
         sopt(fd, TCP_NOPUSH, 1);
     case 3:
         sopt(fd, TCP_NODELAY, 1);
     case 2:
     break;
     }
}

int main(int argc, char const *argv[])
{
    int server_fd, new_socket, valread;
    struct sockaddr_in address;
    int opt = 1;
    int addrlen = sizeof(address);



    // Create socket for TCP server
    server_fd = socket(AF_INET, SOCK_STREAM, 0);

    popts("original", server_fd);
    // Set TCP_NODELAY so that writes won't be batched
    while ((opt = getopt(argc, argv, "sn:o:")) != -1) {
    switch (opt) {
    case 's': step = ! step; break;
    case 'n': nap = strtol(optarg, NULL, 0); break;
    case 'o':
        for (int i = 0; optarg[i]; i++) {
            switch (optarg[i]) {
            case 't': sopt(server_fd, TCP_NODELAY, 1); break;
            case 'p': sopt(server_fd, TCP_NOPUSH, 0); break;
            case 's': sopt(server_fd, SO_SNDBUF, 1); break;
            case 'l': sopt(server_fd, SO_SNDLOWAT, 1); break;
            default:
                exit(1);
            }
        }
    }
    }
    address.sin_family = AF_INET;
    address.sin_addr.s_addr = INADDR_ANY;
    address.sin_port = htons(8080);

    if (bind(server_fd, (struct sockaddr *)&address, sizeof(address)) == -1) {
    xerror("bind");
    }
    popts("ready", server_fd);
    while (1) {
        if (listen(server_fd, 3) == -1) {
        xerror("listen");
        }

        // Accept one client connection
        new_socket = accept(server_fd, (struct sockaddr *)&address, (socklen_t*)&addrlen);
        if (new_socket == -1) {
        xerror("accept");
        }
            popts("accepted: ", new_socket);
        sopt(new_socket, SO_SNDBUF, gopt(server_fd, SO_SNDBUF));
        sopt(new_socket, SO_SNDLOWAT, gopt(server_fd, SO_SNDLOWAT));
        if (step) {
                stepsock(new_socket);
            }
        long tick[21];
        tick[0] = rdtsc();
        // Make N sequential writes to kernel buffer.
        for (int i = 0; i < N; i++) {
                char ch = 'a' + i;

        write(new_socket, &ch, 1);
        tick[i+1] = rdtsc();

        // For some reason, doing some blocking work between the writes
        // The following work slows down the writes by a factor of 10.
        if (nap) {
           sleep(nap);
        }
        }
        for (int i = 1; i < N+1; i++) {
        printf("%ld\n", tick[i] - tick[i-1]);
        }
        printf("_\n");

        // Print a prime to make sure the compiler doesn't optimize
        // away the computations.
        close(new_socket);
    }
}

clnt.c:

#include <stdio.h>
#include <stdlib.h>
#include <netinet/in.h>
#include <netinet/tcp.h>
#include <string.h>
#include <unistd.h>

#ifndef N
#define N 20
#endif
int nap = 0;

int main(int argc, char const *argv[])
{
    int sock, valread;
    struct sockaddr_in address;
    int opt = 1;
    int addrlen = sizeof(address);

    // We'll be passing uint32's back and forth
    unsigned char recv_buffer[1024] = {0};

    // Create socket for TCP server
    sock = socket(AF_INET, SOCK_STREAM, 0);

    // Set TCP_NODELAY so that writes won't be batched
    setsockopt(sock, SOL_SOCKET, TCP_NODELAY, &opt, sizeof(opt));

    while ((opt = getopt(argc,argv,"n:")) != -1) {
        switch (opt) {
        case 'n': nap = strtol(optarg, NULL, 0); break;
        default:
            exit(1);
        }
    }
    opt = 1;
    address.sin_family = AF_INET;
    address.sin_addr.s_addr = INADDR_ANY;
    address.sin_port = htons(8080);

    // Accept one client connection
    if (connect(sock, (struct sockaddr *)&address, (socklen_t)addrlen) != 0) {
        perror("connect failed");
    exit(1);
    }
    if (nap) {
    sleep(nap);
    }
    int loc[N+1];
    int nloc, curloc; 
    for (nloc = curloc = 0; curloc < N; nloc++) {
    int n = read(sock, recv_buffer + curloc, sizeof recv_buffer-curloc);
        if (n <= 0) {
        perror("read");
        break;
    }
    curloc += n;
    loc[nloc] = curloc;
    }
    int last = 0;
    for (int i = 0; i < nloc; i++) {
    int t = loc[i] - last;
    printf("%*.*s ", t, t, recv_buffer + last);
    last = loc[i];
    }
    printf("\n");
    return 0;
}

rdtsc.s:

.globl _rdtsc
_rdtsc:
    rdtsc
    shl $32, %rdx
    or  %rdx,%rax
    ret