• 你知道你的电脑1秒钟能做多少事情吗?(转)


    英文原文:DO YOU KNOW HOW MUCH YOUR COMPUTER CAN DO IN A SECOND?

      让我们来看看你有多么了解电脑!所有这些程序的数值都是可变的。你的任务是:在程序花费 1 秒运行之前猜测它的大概值。

      你并不需要猜出一个精确值:选择范围在 1 和 10 亿之间。你只要能猜出正确的数量级,就算正确!下面是一些注意事项:

    • 如果答案是 38,000,那么你选择 10,000 或 100,000,我们就认为都是正确答案。误差只要在 10 倍范围内就 ok:)
    • 我们知道不同的计算机有不同的磁盘、网络和 CPU 速度!我们会告诉运行 10 次/秒和 10 万次/秒的代码之间的差别。更新的电脑不会让你的代码运行速度快 1000 倍:)
    • 也就是说,所有这一切都是运行在一台新的拥有一个快速的 SSD 和一个凑合的网络连接的笔记本电脑上的。 C 代码用 gcc -O2 编译。

      祝你好运!

    欢迎来到第一个程序!这一个只是让你练练手的:1 秒能完成多少循环? (结果可能比你想象得更多!)

      猜猜下面的程序每秒执行多少次循环:

    #include <stdlib.h>
    
    // Number to guess: How many iterations of
    // this loop can we go through in a second?
    
    int main (int argc, char **argv) {
        int NUMBER, i, s;
        NUMBER = atoi (argv[1]);
    
        for (s = i = 0; i < NUMBER; ++i) {
            s += 1;
        }
    
        return 0;
    }

      准确答案:550,000,000

      猜猜下面的程序每秒执行多少次循环:

    #!/usr/bin/env python
    
    # Number to guess: How many iterations of an
    # empty loop can we go through in a second?
    
    def f (NUMBER):
        for _ in xrange (NUMBER):
            pass
    
    import sys
    f (int(sys.argv[1]))

      准确答案:68,000,000

      当我看着代码的时候,我想的是 1 毫秒完成多少次——我以为是微不足道的,但事实是,即使是 Python,你也可以在 1 毫秒的时间内执行 68,000 次空循环迭代。

    下面让我们来探讨一个更接近现实的用例。在 Python 中字典几乎是无处不在的,那么在 1 秒时间内我们可以用 Python 添加多少元素呢?

    然后再来看一个更复杂的操作——使用 Python 的内置 HTTP 请求解析器来解析请求。

      猜猜下面的程序每秒执行多少次循环:

    #!/usr/bin/env python
    
    # Number to guess: How many entries can
    # we add to a dictionary in a second?
    
    # Note: we take `i % 1000` to control
    # the size of the dictionary
    
    def f (NUMBER):
        d = {}
        for i in xrange (NUMBER):
            d[i % 1000] = i
    
    import sys
    f (int(sys.argv[1]))

      准确答案:11,000,000

      猜猜下面的程序每秒处理多少次 HTTP 请求:

    #!/usr/bin/env python
    
    # Number to guess: How many HTTP requests
    # can we parse in a second?
    
    from BaseHTTPServer import BaseHTTPRequestHandler
    from StringIO import StringIO
    
    class HTTPRequest (BaseHTTPRequestHandler):
        def __init__(self, request_text):
            self.rfile = StringIO (request_text)
            self.raw_requestline = self.rfile.readline ()
            self.error_code = self.error_message = None
            self.parse_request ()
    
        def send_error (self, code, message):
            self.error_code = code
            self.error_message = message
    
    request_text = """GET / HTTP/1.1
    Host: localhost:8001
    Connection: keep-alive
    Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8
    Upgrade-Insecure-Requests: 1
    User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/45.0.2454.85 Safari/537.36
    Accept-Encoding: gzip, deflate, sdch
    Accept-Language: en-GB,en-US;q=0.8,en;q=0.6
    """
    
    def f (NUMBER):
        for _ in range (NUMBER):
            HTTPRequest (request_text)
    
    import sys
    f (int(sys.argv[1]))

      准确答案:25,000

      我们每秒可以解析 25,000 个小的 HTTP 请求!有一件事我要在这里指出的是,这里请求解析的代码是用纯 Python 编写的,而不是C。

    接下来,我们要试试下载网页与运行 Python 脚本!提示:少于 1 亿:)

      猜猜下面的程序每秒可以完成多少次 HTTP 请求:

    #!/usr/bin/env python
    
    # Number to guess: How many times can we
    # download google.com in a second?
    
    from urllib2 import urlopen
    
    def f (NUMBER):
        for _ in xrange (NUMBER):
            r = urlopen ("http://google.com")
            r.read ()
    
    import sys
    f (int(sys.argv[1]))

      准确答案:4

      猜猜下面的程序每秒可以执行多少次循环:

    #!/bin/bash
    
    # Number to guess: How many times can we start
    # the Python interpreter in a second?
    
    NUMBER=$1
    
    for i in $(seq $NUMBER); do
        python -c '';
    done

      准确答案:77

      启动程序实际上昂贵在其本身,而不是启动 Python。如果我们只是运行/bin/true,那么 1 秒能做 500 次,所以看起来运行任何程序只需要大约 1 毫秒时间。当然,下载网页的快慢很大程度上取决于网页大小,网络连接速度,以及服务器间的距离,不过今天我们不谈网络性能。我的一个朋友说,高性能的网络完成网络往返甚至可能只要 250 纳秒(!!!),但这是在计算机位置更相邻,硬件更好的情况下。

    1 秒时间能够在磁盘中写入多少字节?我们都知道写到内存中时速度会更快,但是究竟会快多少呢?对了,下面的代码运行在带有 SSD 的计算机上。

      猜猜下面的程序每秒可以写入多少字节数据:

    #!/usr/bin/env python
    
    # Number to guess: How many bytes can we write
    # to an output file in a second?
    # Note: we make sure everything is sync'd to disk
    # before exiting
    import tempfile
    import os
    
    CHUNK_SIZE = 1000000
    s = "a" * CHUNK_SIZE
    
    def cleanup (f, name):
        f.flush ()
        os.fsync (f.fileno ())
        f.close ()
        try:
            os.remove (name)
        except:
            pass
    
    def f (NUMBER):
        name = './out'
        f = open (name, 'w')
        bytes_written = 0
        while bytes_written < NUMBER:
            f.write (s)
            bytes_written += CHUNK_SIZE
        cleanup (f, name)
    
    import sys
    f (int(sys.argv[1]))

      准确答案:342,000,000

      猜猜下面的程序每秒可以写入多少字节数据:

    #!/usr/bin/env python
    
    # Number to guess: How many bytes can we write
    # to a string in memory in a second?
    
    import cStringIO
    
    CHUNK_SIZE = 1000000
    s = "a" * CHUNK_SIZE
    
    def f (NUMBER):
        output = cStringIO.StringIO ()
        bytes_written = 0
        while bytes_written < NUMBER:
            output.write (s)
            bytes_written += CHUNK_SIZE
    
    import sys
    f (int(sys.argv[1]))

      准确答案:2,000,000,000

    下面轮到文件了!有时候,运行一个大型的 grep 之后,它可以永恒跑下去。在 1 秒时间内,grep 可以搜索多少字节?

    请注意,在这么做的时候,grep 正在读取的字节已经在内存中。

    文件列表同样需要时间!1 秒能列出多少文件?

      猜猜下面的程序每秒可以搜索多少字节的数据:

    #!/bin/bash 
    
    # Number to guess: How many bytes can `grep`
    # search, unsuccessfully, in a second?
    # Note: the bytes are in memory
    
    NUMBER=$1
    
    cat /dev/zero | head -c $NUMBER | grep blah
    exit 0

      准确答案:2,000,000,000

      猜猜下面的程序每秒可以列出多少文件:

    #!/bin/bash
    
    # Number to guess: How many files can `find` list in a second?
    # Note: the files will be in the filesystem cache.
    
    find / -name '*' 2> /dev/null | head -n $1 > /dev/null

      准确答案:325,000

    序列化是一个普遍要花费大量时间的地方,让人很蛋疼,特别是如果你反复结束序列化/反序列化相同数据的时候。这里有几个基准:转换 64K 大小的 JSON 格式数据,与同样大小的 msgpack 格式数据。

      猜猜下面的程序每秒可以执行多少次循环:

    #!/usr/bin/env python
    
    # Number to guess: How many times can we parse
    # 64K of JSON in a second?
    
    import json
    
    with open ('./setup/protobuf/message.json') as f:
        message = f.read ()
    
    def f (NUMBER):
        for _ in xrange (NUMBER):
            json.loads (message)
    
    import sys
    f (int(sys.argv[1]))

      准确答案:449

      猜猜下面的程序每秒可以执行多少次循环:

    #!/usr/bin/env python
    
    # Number to guess: How many times can we parse
    # 46K of msgpack data in a second?
    
    import msgpack
    
    with open ('./setup/protobuf/message.msgpack') as f:
        message = f.read ()
    
    def f (NUMBER):
        for _ in xrange (NUMBER):
            msgpack.unpackb (message)
    
    import sys
    f (int(sys.argv[1]))

      准确答案:4,000

    数据库。没有任何类似于 PostgreSQL 花里胡哨的东西,我们做了 2 份有 1000 万行数据的 SQLite 表,一个是有索引的,另一个是未建索引的。

      猜猜下面的程序每秒可以执行多少次查询:

    #!/usr/bin/env python
    
    # Number to guess: How many times can we
    # select a row from an **indexed** table with 
    # 10,000,000 rows?
    
    import sqlite3
    
    conn = sqlite3.connect ('./indexed_db.sqlite')
    c = conn.cursor ()
    def f (NUMBER):
        query = "select * from my_table where key = %d" % 5
        for i in xrange (NUMBER):
            c.execute (query)
            c.fetchall ()
    
    import sys
    f (int(sys.argv[1]))

      准确答案:53,000

      猜猜下面的程序每秒执行多少次查询:

    #!/usr/bin/env python
    
    # Number to guess: How many times can we
    # select a row from an **unindexed** table with 
    # 10,000,000 rows?
    
    import sqlite3
    
    conn = sqlite3.connect ('./unindexed_db.sqlite')
    c = conn.cursor ()
    def f (NUMBER):
        query = "select * from my_table where key = %d" % 5
        for i in xrange (NUMBER):
            c.execute (query)
            c.fetchall ()
    
    import sys
    f (int(sys.argv[1]))

      准确答案:2

    下面要说 Hash 算法!在这里,我们将比较 MD5 和 bcrypt。用 MD5 你在 1 秒时间内可以哈希到相当多的东西,而用 bcrypt 则不能。

      猜猜下面的程序每秒可以哈希多少字节的数据:

    #!/usr/bin/env python
    
    # Number to guess: How many bytes can we md5sum in a second?
    
    import hashlib
    
    CHUNK_SIZE = 10000
    s = 'a' * CHUNK_SIZE
    
    def f (NUMBER):
        bytes_hashed = 0
        h = hashlib.md5()
        while bytes_hashed < NUMBER:
            h.update (s)
            bytes_hashed += CHUNK_SIZE
        h.digest ()
    import sys
    f (int(sys.argv[1]))

      准确答案:455,000,000

      猜猜下面的程序每秒可以哈希多少字节的密码:

    #!/usr/bin/env python
    
    # Number to guess: How many passwords
    # can we bcrypt in a second?
    
    import bcrypt
    
    password = 'a' * 100
    
    def f (NUMBER):
        for _ in xrange (NUMBER):
            bcrypt.hashpw (password, bcrypt.gensalt ())
    
    import sys
    f (int(sys.argv[1]))

      准确答案:3

    接下来,我们要说一说内存访问。 现在的 CPU 有 L1 和 L2 缓存,这比主内存访问速度更快。这意味着,循序访问内存通常比不按顺序访问内存能提供更快的代码。

      猜猜下面的程序每秒可以向内存写入多少字节数据:

    #include <stdlib.h>
    #include <stdio.h>
    
    // Number to guess: How big of an array (in bytes)
    // can we allocate and fill in a second?
    
    // this is intentionally more complicated than it needs to be
    // so that it matches the out-of-order version
    
    int main (int argc, char **argv) {
        int NUMBER, i;
        NUMBER = atoi (argv[1]);
    
        char* array = malloc (NUMBER);
        int j = 1;
        for (i = 0; i < NUMBER; ++i) {
            j = j * 2;
            if (j > NUMBER) {
                j = j - NUMBER;
            }
            array[i] = j;
        }
    
        printf ("%d", array[NUMBER / 7]);
        // so that -O2 doesn't optimize out the loop
    
        return 0;
    }

      准确答案:376,000,000

      猜猜下面的程序每秒可以向内存写入多少字节数据:

    #include <stdlib.h>
    #include <stdio.h>
    
    // Number to guess: How big of an array (in bytes)
    // can we allocate and fill with 5s in a second?
    // The catch: We do it out of order instead of in order.
    int main (int argc, char **argv) {
        int NUMBER, i;
        NUMBER = atoi (argv[1]);
    
        char* array = malloc (NUMBER);
        int j = 1;
        for (i = 0; i < NUMBER; ++i) {
            j = j * 2;
            if (j > NUMBER) {
                j = j - NUMBER;
            }
            array[j] = j;
        }
    
        printf ("%d", array[NUMBER / 7]);
        // so that -O2 doesn't optimize out the loop
    
        return 0;
    }

      准确答案:68,000,000

      欢迎大家去试一试,给我们留下宝贵的意见。

      -

      译文链接:http://www.codeceo.com/article/1-second-your-computer-do.html

  • 相关阅读:
    Nginx实战系列之功能篇----后端节点健康检查
    nginx大量TIME_WAIT的解决办法
    Nginx 获取真实 IP 方案
    Redis基本操作——List
    redis-cli 命令总结
    redis 学习笔记-cluster集群搭建
    redis集群部署及常用的操作命令_01
    redis配置文件参数详解
    Redis 主从配置和参数详解
    Tomcat源码分析-开篇(Tomcat源码部署运行 Maven方式)
  • 原文地址:https://www.cnblogs.com/softidea/p/4935590.html
Copyright © 2020-2023  润新知