Default ip is not avaliable. Please make sure the default ip is shown in ip addr, or use a customer ip instead.

时间: 2024-06-06 14:06:59 浏览: 235
As an AI language model, I cannot perform the task of checking or changing the default IP address of a device or system. However, here are some general tips that may help: First, check if the default IP address is correctly shown in the output of the "ip addr" command. If not, you may need to configure the network settings of the device or system to set the correct IP address. If you are unable to find the default IP address or cannot change it, you can try using a custom IP address instead. Make sure the custom IP address is valid for the network you are connecting to and does not conflict with other devices on the same network. If you are still having trouble, you may need to consult the documentation or support resources for your specific device or system for further assistance.
阅读全文

相关推荐

zip

vga.zip_All in the Timing_Quartus 12.1_dac vhdl_frequency contro

This details a VGA controller component that handles VGA signal timing, written in VHDL for use with CPLDs and FPGAs. Figure 1 illustrates a typical example of the VGA controller integrated into a ...
rar

ep.rar_This Is It_huffman

描述中的"values is shown in Fig 19. According to this figure, it can"似乎是在指一个图（图19）展示了某些值，可能是一些符号的频率或者Huffman编码的结果。通常在Huffman编码过程中，我们会根据符号的频率构建...
rar

tennis_game.rar_The Handle_link.r

It is a tennis game made in JAVA. Two player can play tennis. by using keyboard they can handle the slider that shown on the both side of the Frame.

(Displaying a TicTacToe board) Display a frame that contains nine labels. A labelmay display a cross image icon, a not image icon, or nothing, as shown in Figure 12.18(a). What to display is randomly decided. Use the Math.random()method to generate an integer0,1, or 2, which corresponds to displaying across image icon, a not image icon, or nothing. The cross and not imagescanbe obtained from the cross.gif and not.gif files in the image directory onthe Companion Website.

ImageIcon icon = new ImageIcon("image/not.gif"); // 创建一个显示not.gif的图标对象 label.setIcon(icon); // 设置标签的图标为not.gif } // 其它情况标签不显示图标 add(label); // 将标签添加到窗口中 }...

The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: Expression: ( V | V ) & F & ( F | V ) where V is for True, and F is for False. The expressions may include the following operators: ! for not, & for and, | for or, the use of parenthesis for operations grouping is also allowed. To perform the evaluation of an expression, it will be considered the priority of the operators, the not having the highest, and the or the lowest. The program must yield V or F, as the result for each expression in the input file. Input The expressions are of a variable length, although will never exceed 100 symbols. Symbols may be separated by any number of spaces or no spaces at all, therefore, the total length of an expression, as a number of characters, is unknown. The number of expressions in the input file is variable and will never be greater than 20. Each expression is presented in a new line, as shown below. Output For each test expression, print "Expression " followed by its sequence number, ": ", and the resulting value of the corresponding test expression. Separate the output for consecutive test expressions with a new line. Use the same format as that shown in the sample output shown below. Sample Input ( V | V ) & F & ( F| V) !V | V & V & !F & (F | V ) & (!F | F | !V & V) (F&F|V|!V&!F&!(F|F&V)) Sample Output Expression 1: F Expression 2: V Expression 3: V

Note that the evaluate() function uses a stack to keep track of the operands and operators in the expression. It starts by iterating over each symbol in the expression, and for each symbol, it ...

请用python语言实现以下问题： 要求不用evaluate函数，改用栈实现这个问题 The objective of the program you are going to produce is to evaluate boolean expressions as the one shown next: Expression: ( V | V ) & F & ( F | V )where V is for True, and F is for False. The expressions may include the following operators: ! for not, & for and, | for or, the use of parenthesis for operations grouping is also allowed. To perform the evaluation of an expression, it will be considered the priority of the operators, the not having the highest, and the or the lowest. The program must yield V or F, as the result for each expression in the input file. Input The expressions are of a variable length, although will never exceed 100 symbols. Symbols may be separated by any number of spaces or no spaces at all, therefore, the total length of an expression, as a number of characters, is unknown. The number of expressions in the input file is variable and will never be greater than 20. Each expression is presented in a new line, as shown below. Output For each test expression, print "Expression " followed by its sequence number, ": ", and the resulting value of the corresponding test expression. Separate the output for consecutive test expressions with a new line. Use the same format as that shown in the sample output shown below. Sample Input ( V | V ) & F & ( F| V) !V | V & V & !F & (F | V ) & (!F | F | !V & V) (F&F|V|!V&!F&!(F|F&V)) Sample Output Expression 1: F Expression 2: V Expression 3: V

stack.append(not operand) else: # 如果是二元运算符，弹出两个操作数进行运算，并将结果入栈 operand2 = stack.pop() operand1 = stack.pop() if token == '&': stack.append(operand1 and operand2) elif...

A car rental company rents out a variety of cars and coaches and its rental rates are calculated on a daily basis. The relevant information of vehicle is shown in the following form. You need to develop a vehicle rental system according to the following requirements. (1) Design a simple console interactive interface to provide users with the related information about this vehicle rental system including vehicle models, specific information, daily rent and discount information so that the user can make further choices. (2) After understanding the relevant information, the user can input the relevant information of the vehicle which he or she wants to rent through the console program. (3) The program calculates the rent that the user needs to pay and informs the user through the console program. (4) You need to deal with the illegal information input by users through the console with some rules and prompt the corresponding error reason.For example, the user cannot input a negative value for the number of parking days.You need to consider as many misinput cases as possible. Please submit program code and the screenshot of the program output in the answer

To develop a vehicle rental system, you can use a programming language like Python and create a console-based interface using the built-in input and print functions. Here is a general outline of how ...

7 errors have detailed information that is not shown. Use 'stats.errorDetails: true' resp. '--stats-error-details' to show it. webpack 5.83.1 compiled with 11 errors in 1777 ms

这个错误提示是 webpack 在编译过程中出现了 11 个错误，但是并没有给出详细的错误信息，建议你可以加上 --stats-error-details 参数重新编译，以显示详细的错误信息。 你可以在命令行中输入以下命令： ...

Your computer has a keyboard with three keys: 'a' key, Shift key, and Caps Lock key. The Caps Lock key has a light on it. Initially, the light on the Caps Lock key is off, and the screen shows an empty string. You can do the following three actions any number of times in any order: Spend X milliseconds to press only the 'a' key. If the light on the Caps Lock key is off, a is appended to the string on the screen; if it is on, A is. Spend Y milliseconds to press the 'a' key and Shift key simultaneously. If the light on the Caps Lock key is off, A is appended to the string on the screen; if it is on, a is. Spend Z milliseconds to press the Caps Lock key. If the light on the Caps Lock key is off, it turns on; if it is on, it turns off. Given a string S consisting of A and a, determine at least how many milliseconds you need to spend to make the string shown on the screen equal to S.c++代码

Here's a possible C++ implementation of the algorithm I described earlier: #include #include using namespace std; int main() { int X, Y, Z; string S; // read input cin >> X >> Y >> Z >> ...

There is a light sensor on your picoprobe development board. This question asks you to use this sensor to measure the ambient light and to control the three LEDs on the picoprobe board to make a night light. Configure your program to perform the following steps, after printing the message from Question 1. The steps should be repeated every 1 second. raw ADC result in the format shown below: Light Sensor: Raw: _____ (insert the actual value) (15 marks) (b) Convert the raw value to a voltage and print out the voltage that you calculated in the format displayed below. At least one decimal place is required. Voltage: _____ V (insert the actual value) (10 marks) (c) Use this voltage to control the three-colour LED to show a white colour, when the light sensor is covered by a piece of paper (e.g. when the room is getting dark). The white-coloured light will go off, as soon as the ambient light is high again, i.e. when the light sensor is exposed to normal room light. This is reminiscent of an automatic night light.

Sure, here's how you can use the light sensor on the picoprobe development board to create a night light: (a) 1. Initialize the ADC to read from the light sensor. 2. Read the raw ADC value from the ...

下面是一道java编程题In the well-known story of "The tortoise and the hare", the hare was too confident and took a break in the middle of the competition, which led to the tortoise winning the competition. A multi-threaded program is required to simulate the race between the tortoise and the hare. The speed of the tortoise is 1 meter per 1500 milliseconds, and the speed of the hare is 5 meters per 500 milliseconds. When the hare reached the 700th meter, it chose to rest for 10,000 milliseconds. As a result, the tortoise wins the race. The output result of the program running is shown in the figure.

Sure, here is a possible implementation of the program in Java: java public class TortoiseAndHareRace { private static final int RACE_LENGTH = 1000; private static final int TORTOISE_SPEED = 1; ...

Using an ACL connection, a Central can establish one or more isochronous connections that use the isochronous physical channel. An isochronous connection is used to transfer isochronous data between the Central and a Peripheral by using a logical transport, which is referred to as a Connected Isochronous Stream (CIS). A CIS consists of CIS events that occur at regular intervals (designated ISO_Interval). Every CIS event consists of one or more subevents. In each subevent, the Central transmits once and the Peripheral responds. If the Central and Peripheral have completed transferring the scheduled isochronous data in a CIS event, all remaining subevents in that event will have no radio transmissions and the event is closed. Each subevent uses a PHY channel which is determined by using the channel selection algorithm. The PHY channel that is used for a subevent is marked as ISO Ch(eventcount, subeventcount), as shown in Figure 1.4.

通过ACL连接，中央设备可以建立一个或多个使用同步物理通道的同步连接。同步连接用于通过逻辑传输（称为连接的同步流，CIS）在中央设备和外围设备之间传输同步数据。CIS由定期发生的CIS事件组成（指定为ISO_Interval...

1.Consider an employee database as shown in figure 4.12. Write a query to find the ID of an employee who does not have a manager. Note that the employee who will not have manager will be an employee ID that is not present in the manages table.

Sure, here's a SQL query to find the ID of an employee who does not have a manager: SELECT e.ID FROM employee e LEFT JOIN manages m ON e.ID = m.employeeID WHERE m.managerID IS NULL; ...

d:\vbcentos7>vagrant up Bringing machine 'default' up with 'virtualbox' provider... ==> default: Clearing any previously set forwarded ports... ==> default: Clearing any previously set network interfaces... ==> default: Preparing network interfaces based on configuration... default: Adapter 1: nat default: Adapter 2: hostonly ==> default: Forwarding ports... default: 22 (guest) => 2222 (host) (adapter 1) ==> default: Booting VM... There was an error while executing VBoxManage, a CLI used by Vagrant for controlling VirtualBox. The command and stderr is shown below. Command: ["startvm", "2e091b45-6af1-43e0-ae0f-a59caad94d44", "--type", "headless"] Stderr: VBoxManage.exe: error: The VM session was closed before any attempt to power it on VBoxManage.exe: error: Details: code E_FAIL (0x80004005), component SessionMachine, interface ISession

根据你提供的信息，Vagrant 在启动 VirtualBox 虚拟机时遇到了错误。错误信息中提到了 VM session was closed before any attempt to power it on，可能是因为 VirtualBox 无法正常启动虚拟机。...

A Circleobject consists ofa radius. When adding a height to a circle object, a Cylinderobject can be made. A Cylinderobject consists ofa radiusand a height. A default circlecontains a radiusof 1. A default cylindercontains a radiusof 1 and a heightof 1. Assume that both radiusand heightare non-negative integers.Designand implementCircleclass and Cylinderclass. Class Circlemust be the superclass of class Cylinder. Class Circlecannothave any superclass except the ultimate superclass Object. When instantiating the two classes, circles and cylinders can be made and stored in an array list as shown below. publicstaticvoidmain(String[] args){ArrayList<Circle> shapes = newArrayList<Circle>();//Creates a list of circles and cylinders.shapes.add(newCircle());shapes.add(newCircle(5));shapes.add(newCylinder());shapes.add(newCylinder(10,5));//Prints the list.for(Circle next: shapes){System.out.println(next);} }a.Draw an UML diagram to show the design. Visibility modifiers, void/value return type, names, and parameter types are required. (10 points)b.Implement both classes by including only the following members in each of the classes.(35 points each class. 70pointsin total)•instance variables•overloading constructors•getters/setters•overriding toString•overriding equals You must use thisreference and superreference when appropriate. Correct Java syntax is required. No Javadoc style comments are required.

a. 下面是 Circle 类和 Cylinder 类的 UML 图： +--------------------------------+ | Circle | +--------------------------------+ | - radius: int | +--------------------------------+ | + Circle() | |...

use python to finish this task.please show me the code 1) Replicate the same numerical experiments as the examples for pricing barrier option in the PPTs.

Sure, here is the Python code to replicate the numerical experiments for pricing barrier options: python import numpy as np from scipy.stats import norm # Parameters S0 = 100.0 # initial stock ...

A device can use an isochronous physical channel to broadcast isochronous data by using isochronous connectionless logical transports. An isochronous connectionless logical transport is referred to as a Broadcast Isochronous Stream (BIS). A BIS consists of BIS events that occur at regular intervals (designated ISO_Interval). Every BIS event consists of one or more subevents. In every subevent, a broadcasting device transmits an isochronous data packet. Each subevent uses a PHY channel that is determined using the channel selection algorithm. A device can transmit several BISes with synchronized timing; this is referred to as a Broadcast Isochronous Group (BIG). The various BIS events together form a BIG event. The device can also use the isochronous physical channel to broadcast control information in a Control subevent, which is transmitted at the end of all subevents for a BISBIG, as shown in Figure 1.5.

设备可以使用同步物理通道通过使用同步无连接逻辑传输来广播同步数据。同步无连接逻辑传输称为广播同步流（BIS）。BIS由定期发生的BIS事件组成（指定为ISO_Interval）。每个BIS事件由一个或多个子事件组成。...

用C++写You are given two integers x and k . Grasshopper starts in a point 0 on an OX axis. In one move, it can jump some integer distance, that is not divisible by k , to the left or to the right. What's the smallest number of moves it takes the grasshopper to reach point x ? What are these moves? If there are multiple answers, print any of them. Input The first line contains a single integer t (1≤t≤1000 ) — the number of testcases. The only line of each testcase contains two integers x and k (1≤x≤100 ; 2≤k≤100 ) — the endpoint and the constraint on the jumps, respectively. Output For each testcase, in the first line, print a single integer n — the smallest number of moves it takes the grasshopper to reach point x . In the second line, print n integers, each of them not divisible by k . A positive integer would mean jumping to the right, a negative integer would mean jumping to the left. The endpoint after the jumps should be exactly x . Each jump distance should be from −109 to 109 . In can be shown that, for any solution with the smallest number of jumps, there exists a solution with the same number of jumps such that each jump is from −109 to 109 . It can be shown that the answer always exists under the given constraints. If there are multiple answers, print any of them.

以下是本题的 C++ 代码实现： c++ #include #include using namespace std; int main() ...接着根据上文中提到的算法思想进行判断，如果 $x$ 能被 $k$ 整除，则至少需要跳两步，输出 $k$ 和 $x$；...

翻译：As shown in Fig. 2, to solve problems (7) and (11), a multileader multi-follower Stackelberg game SL, in which BSs are regarded as leaders and UE are followers, is adopted. Although players make their decisions independently, a hierarchy of players at different levels exists in this game, and the strategies chosen by the followers depend on the strategies selected by the leaders. Moreover, the choice of the optimal strategy of leaders can vary with the follower equilibrium that occurs. Here, the Stackelberg equilibrium (SE) of the proposed Stackelberg game is defined as follows.

如图2所示，为了解决问题（7）和（11），采用了一个多领导者多追随者斯塔克伯格博弈SL，其中基站被视为领导者，用户设备（UE）则是追随者。尽管玩家们独立地做出决策，但在这个博弈中存在不同层级的玩家，并且追随者...

a page or script is accessing at least one of navigator.useragent, navigator.appversion, and navigator.platform. starting in chrome 101, the amount of information available in the user agent string will be reduced. to fix this issue, replace the usage of navigator.useragent, navigator.appversion, and navigator.platform with feature detection, progressive enhancement, or migrate to navigator.useragentdata. note that for performance reasons, only the first access to one of the properties is shown.

这是Chrome浏览器中的一个提示，指出某个页面或脚本正在访问navigator.useragent、navigator.appversion和navigator.platform等属性。从Chrome 101开始，用户代理字符串中可用信息的数量将被减少。...

最新推荐

recommend-type

Pytorch版代码幻灯片.zip

Jupyter-Notebook
recommend-type

Jupyter_Chat甄嬛是利用甄嬛传剧本中所有关于甄嬛的台词和语句基于ChatGLM2进行LoRA微调得到的模仿甄.zip

Jupyter-Notebook
recommend-type

高效甘特图模板下载-精心整理.zip

高效甘特图模板下载-精心整理.zip
recommend-type

高清艺术文字图标资源，PNG和ICO格式免费下载

资源摘要信息:"艺术文字图标下载" 1. 资源类型及格式：本资源为艺术文字图标下载，包含的图标格式有PNG和ICO两种。PNG格式的图标具有高度的透明度以及较好的压缩率，常用于网络图形设计，支持24位颜色和8位alpha透明度，是一种无损压缩的位图图形格式。ICO格式则是Windows操作系统中常见的图标文件格式，可以包含不同大小和颜色深度的图标，通常用于桌面图标和程序的快捷方式。 2. 图标尺寸：所下载的图标尺寸为128x128像素，这是一个标准的图标尺寸，适用于多种应用场景，包括网页设计、软件界面、图标库等。在设计上，128x128像素提供了足够的面积来展现细节，而大尺寸图标也可以方便地进行缩放以适应不同分辨率的显示需求。 3. 下载数量及内容：资源提供了12张艺术文字图标。这些图标可以用于个人项目或商业用途，具体使用时需查看艺术家或资源提供方的版权声明及使用许可。在设计上，艺术文字图标融合了艺术与文字的元素，通常具有一定的艺术风格和创意，使得图标不仅具备标识功能，同时也具有观赏价值。 4. 设计风格与用途：艺术文字图标往往具有独特的设计风格，可能包括手绘风格、抽象艺术风格、像素艺术风格等。它们可以用于各种项目中，如网站设计、移动应用、图标集、软件界面等。艺术文字图标集可以在视觉上增加内容的吸引力，为用户提供直观且富有美感的视觉体验。 5. 使用指南与版权说明：在使用这些艺术文字图标时，用户应当仔细阅读下载页面上的版权声明及使用指南，了解是否允许修改图标、是否可以用于商业用途等。一些资源提供方可能要求在使用图标时保留作者信息或者在产品中适当展示图标来源。未经允许使用图标可能会引起版权纠纷。 6. 压缩文件的提取：下载得到的资源为压缩文件，文件名称为“8068”，意味着用户需要将文件解压缩以获取里面的PNG和ICO格式图标。解压缩工具常见的有WinRAR、7-Zip等，用户可以使用这些工具来提取文件。 7. 具体应用场景：艺术文字图标下载可以广泛应用于网页设计中的按钮、信息图、广告、社交媒体图像等；在应用程序中可以作为启动图标、功能按钮、导航元素等。由于它们的尺寸较大且具有艺术性，因此也可以用于打印材料如宣传册、海报、名片等。 通过上述对艺术文字图标下载资源的详细解析，我们可以看到，这些图标不仅是简单的图形文件，它们集合了设计美学和实用功能，能够为各种数字产品和视觉传达带来创新和美感。在使用这些资源时，应遵循相应的版权规则，确保合法使用，同时也要注重在设计时根据项目需求对图标进行适当调整和优化，以获得最佳的视觉效果。
recommend-type

管理建模和仿真的文件

管理Boualem Benatallah引用此版本：布阿利姆·贝纳塔拉。管理建模和仿真。约瑟夫-傅立叶大学-格勒诺布尔第一大学，1996年。法语。NNT：电话：00345357HAL ID：电话：00345357https://theses.hal.science/tel-003453572008年12月9日提交HAL是一个多学科的开放存取档案馆，用于存放和传播科学研究论文，无论它们是否被公开。论文可以来自法国或国外的教学和研究机构，也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaire
recommend-type

DMA技术：绕过CPU实现高效数据传输

![DMA技术：绕过CPU实现高效数据传输](https://res.cloudinary.com/witspry/image/upload/witscad/public/content/courses/computer-architecture/dmac-functional-components.png) # 1. DMA技术概述 DMA（直接内存访问）技术是现代计算机架构中的关键组成部分，它允许外围设备直接与系统内存交换数据，而无需CPU的干预。这种方法极大地减少了CPU处理I/O操作的负担，并提高了数据传输效率。在本章中，我们将对DMA技术的基本概念、历史发展和应用领域进行概述，为读
recommend-type

SGM8701电压比较器如何在低功耗电池供电系统中实现高效率运作？

SGM8701电压比较器的超低功耗特性是其在电池供电系统中高效率运作的关键。其在1.4V电压下工作电流仅为300nA，这种低功耗水平极大地延长了电池的使用寿命，尤其适用于功耗敏感的物联网(IoT)设备，如远程传感器节点。SGM8701的低功耗设计得益于其优化的CMOS输入和内部电路，即使在电池供电的设备中也能提供持续且稳定的性能。 参考资源链接：[SGM8701：1.4V低功耗单通道电压比较器](https://wenku.csdn.net/doc/2g6edb5gf4?spm=1055.2569.3001.10343) 除此之外，SGM8701的宽电源电压范围支持从1.4V至5.5V的电
recommend-type

mui框架HTML5应用界面组件使用示例教程

资源摘要信息:"HTML5基本类模块V1.46例子（mui角标+按钮+信息框+进度条+表单演示）-易语言" 描述中的知识点: 1. HTML5基础知识：HTML5是最新一代的超文本标记语言，用于构建和呈现网页内容。它提供了丰富的功能，如本地存储、多媒体内容嵌入、离线应用支持等。HTML5的引入使得网页应用可以更加丰富和交互性更强。 2. mui框架：mui是一个轻量级的前端框架，主要用于开发移动应用。它基于HTML5和JavaScript构建，能够帮助开发者快速创建跨平台的移动应用界面。mui框架的使用可以使得开发者不必深入了解底层技术细节，就能够创建出美观且功能丰富的移动应用。 3. 角标+按钮+信息框+进度条+表单元素：在mui框架中，角标通常用于指示未读消息的数量，按钮用于触发事件或进行用户交互，信息框用于显示临时消息或确认对话框，进度条展示任务的完成进度，而表单则是收集用户输入信息的界面组件。这些都是Web开发中常见的界面元素，mui框架提供了一套易于使用和自定义的组件实现这些功能。 4. 易语言的使用：易语言是一种简化的编程语言，主要面向中文用户。它以中文作为编程语言关键字，降低了编程的学习门槛，使得编程更加亲民化。在这个例子中，易语言被用来演示mui框架的封装和使用，虽然描述中提到“如何封装成APP，那等我以后再说”，暗示了mui框架与移动应用打包的进一步知识，但当前内容聚焦于展示HTML5和mui框架结合使用来创建网页应用界面的实例。 5. 界面美化源码：文件的标签提到了“界面美化源码”，这说明文件中包含了用于美化界面的代码示例。这可能包括CSS样式表、JavaScript脚本或HTML结构的改进，目的是为了提高用户界面的吸引力和用户体验。 压缩包子文件的文件名称列表中的知识点: 1. mui表单演示.e：这部分文件可能包含了mui框架中的表单组件演示代码，展示了如何使用mui框架来构建和美化表单。表单通常包含输入字段、标签、按钮和其他控件，用于收集和提交用户数据。 2. mui角标+按钮+信息框演示.e：这部分文件可能展示了mui框架中如何实现角标、按钮和信息框组件，并进行相应的事件处理和样式定制。这些组件对于提升用户交互体验至关重要。 3. mui进度条演示.e：文件名表明该文件演示了mui框架中的进度条组件，该组件用于向用户展示操作或数据处理的进度。进度条组件可以增强用户对系统性能和响应时间的感知。 4. html5标准类1.46.ec：这个文件可能是核心的HTML5类库文件，其中包含了HTML5的基础结构和类定义。"1.46"表明这是特定版本的类库文件，而".ec"文件扩展名可能是易语言项目中的特定格式。 总结来说，这个资源摘要信息涉及到HTML5的前端开发、mui框架的界面元素实现和美化、易语言在Web开发中的应用，以及如何利用这些技术创建功能丰富的移动应用界面。通过这些文件和描述，可以学习到如何利用mui框架实现常见的Web界面元素，并通过易语言将这些界面元素封装成移动应用。
recommend-type

"互动学习：行动中的多样性与论文攻读经历"

多样性她- 事实上SCI NCES你的时间表ECOLEDO C Tora SC和NCESPOUR l’Ingén学习互动，互动学习以行动为中心的强化学习学会互动，互动学习，以行动为中心的强化学习计算机科学博士论文于2021年9月28日在Villeneuve d'Asq公开支持马修·瑟林评审团主席法布里斯·勒菲弗尔阿维尼翁大学教授论文指导奥利维尔·皮耶昆谷歌研究教授：智囊团论文联合主任菲利普·普雷教授，大学。里尔/CRISTAL/因里亚报告员奥利维耶·西格德索邦大学报告员卢多维奇·德诺耶教授，Facebook /索邦大学审查员越南圣迈IMT Atlantic高级讲师邀请弗洛里安·斯特鲁布博士，Deepmind对于那些及时看到自己错误的人...3谢谢你首先，我要感谢我的两位博士生导师Olivier和Philippe。奥利维尔，"站在巨人的肩膀上"这句话对你来说完全有意义了。从科学上讲，你知道在这篇论文的（许多）错误中，你是我可以依
recommend-type

【数据传输高速公路】：总线系统的深度解析

![计算机组成原理知识点](https://img-blog.csdnimg.cn/6ed523f010d14cbba57c19025a1d45f9.png) # 1. 总线系统概述 在计算机系统和电子设备中，总线系统扮演着至关重要的角色。它是一个共享的传输介质，用于在组件之间传递数据和控制信号。无论是单个芯片内部的互连，还是不同设备之间的通信，总线技术都是不可或缺的。为了实现高效率和良好的性能，总线系统必须具备高速传输能力、高效的数据处理能力和较高的可靠性。 本章节旨在为读者提供总线系统的初步了解，包括其定义、历史发展、以及它在现代计算机系统中的应用。我们将讨论总线系统的功能和它在不同层