2011年12月1日

随着Scrum的引入，测试被提到很重要的地位，特别是自动化测试, UI Automation是其中一个重要的部分，用来模拟用户的操作，来发现可能存在的问题，除了测试用例的有效，质量，速度也是必须考虑的。现在有这么一个场景，如果你有时间，并且感兴趣的话，不妨试一试，如果你的总体运行时间小于或者等于2秒的话，请说出你的实现逻辑，使用的工具，机器的相关配置。

场景：在记事本里面输入"hello World",并将其保存在C:\a.txt，如果存在则替换。

步骤：

1. 打开Notepad.exe
2. 输入hello world
3. 在工具栏点击Close 按钮
4. 在确认对话框中选择"保存"
5. 输入文件路径和名字　C:\a.txt
6. 如果文件已经存在，在确认对话框中选择"是

  2011年4月27日

Configuration Management

Version Control

• Keep absolutely everything in Version Control

• Check in regularly to Trunk

• Use meaningful commit messages
• Managing Dependencies
• Managing external libraries

• Managing components

• Managing software configuration

• Configuration and flexibility

• Type of Configuration
• Build time
• Packaging time
• Deployment time
• Startup time or run time

• Managing Application Configuration
• Accessing Configuration
• Modeling Configuration
• Adding
• Creating
• Promoting
• Relocating
• Managing

• Testing System Configuration
• Managing configuration across applications

Principles of managing application configuration

Managing your environments

• Various operating systems, version, patch levels, configuration settings

• Additional software packages, version, configuration

• Networking topology

• External services

• Any Data or other state

Tools to manage environments

Manage the change process

What's the foundation of continuous delivery?

it's just Automation+Delivery.

Every Change Should Trigger the feedback process

• Executable Code
• Configuration

• Host Environment

• Data
  • Unit Tests
  • Test Coverage and other technology specific metrics
  • Conform to tis business acceptance criteria
  • Non-function (Capacity, Availability, Security)
  • Exploratory Testing

Principles of software delivery

• Create a repeatable, reliable process for releasing software
• Automate almost everything
• Keep everything in version control
• If it hurts, do it more frequently, and bring the pain forward (Heuristic)
• Build Quality In
• Done means Released

• Everybody is responsible for the delivery process

• Continuous Improvement

  2011年2月15日

CAB, in terms of Composite Application block, published by Microsoft Pattern & Design team. today we try to learn the event aggregator design pattern included in CAB, which could help understand how it's designed and implemented.

As unusal, module would interact with other module or itself, so it needs to take a reference other module, couple-loosed. is there decouple-loose? The anser is true,Event Aggregator,which could resolve this problem, there three important parts, Publisher, Subscriber and EventAggregator.

• Publisher, it just needs to publish its event and doesn't know the specific subscriber, and how many subscribers there are.
• Subscriber, justs subscribe its care-about event, and doesn't know who's the publisher.
• EventAggregator, just a transfer station, used to store all the subscribers for specific event. when publisher triggers its event, it would find all matched event to execute, within CAB, there are three different execution style.
• Background Thread
• UIThread 
• Executing Thread

IEvent Interface

 1     public interface IEvent<TPayload>
 2     {
 3         SubscriptionToken Subscribe(Action<TPayload> action);
 4         SubscriptionToken Subscribe(Action<TPayload> action, ThreadOption threadOption);
 5         SubscriptionToken Subscribe(Action<TPayload> action, bool keepSubscriberReferenceAlive);
 6         SubscriptionToken Subscribe(Action<TPayload> action, ThreadOption threadOption, bool keepSubscriberReferenceAlive);
 7         SubscriptionToken Subscribe(Action<TPayload> action, ThreadOption threadOption, bool keepSubscriberReferenceAlive, Predicate<TPayload> filter);
 8         void Publish(TPayload payload);
 9         void Unsubscribe(SubscriptionToken token);
10         void Unsubscribe(Action<TPayload> subscriber);
11         bool Contains(SubscriptionToken token);
12         bool Contains(Action<TPayload> subscriber);
13     }

  2011年1月28日

如果寻找给定的数组中第几大的数,采用前面的方法就不现实,因为我们不可能定义几个变量来存放,并且也不现实,这时我们当然想到了数组,而且是有序数组,按照从大到小的方式进行排序,再套用前面的逻辑就可以了.

代码

        internal static int FindIndexMaximum(int[] input, int index)
        {
            if (input == null || input.Length < index)
            {
                throw new InvalidOperationException("the input paramter is invalid, please check your input");
            }
            //Get the sorted array
            int[] sorted = new int[index];
            int k = 0;
            for (int i = 0; i < index; i++)
            {
                k = i;
                while (k > 0 && input[i] > sorted[k - 1])
                {
                    sorted[k] = sorted[k - 1];
                    k--;
                }
                sorted[k] = input[i];
            }
            //Compare the number one by one.
            for (int i = index; i < input.Length; i++)
            {
                k = index;
                while (k > 0 && input[i] > sorted[k - 1])
                {
                    if (k != index)
                    {
                        sorted[k] = sorted[k - 1];
                    }
                    k--;
                }
                if (input[i] > sorted[index - 1])
                {
                    sorted[k] = input[i];
                }
            }
            return sorted[index - 1];
        }

其实这里主要是使用了两次插入式排序方式,第一次是用来初始化临时数组,并保存其有序.第二次就是从原始数组的index位置开始取数,一个一个地与临时数组的最小值进行比较,如果小于最小值就跳过,如果大于,则一一比较,并插入到合适的位置.这样临时数组的最后一个(?)就是你要找的数字. 这儿还是有一个问题在这儿,如果前面的数字有重复的,那这儿就会有问题?如果让你来改进,你会如何做呢?

另外,让你来设计它的测试用例,你会给出哪些呢? 如果你有不错的想法,请与大家分享.

1. Logic and visual trees

• What's the difference between them
• How to use them in effective

2. Dpendency Properites

• Enable styling
• Automatic data binding
• Animation
• ...

what's the problem that dependency tries to resolve?

• Change notification
• property value inheritance
• support for multiple providers

Dependency property implementation

Trigger

• Property triggers
• Data Triggers
• Event Triggers

3.Routed Events

4.Commands

  2011年1月27日

给定一个数组,寻找次大的数,你会如何实现?

代码

 internal static int FindSecondMaximum(int[] input)
        {
            if (input == null || input.Length < 2)
            {
                throw new InvalidOperationException("the input paramter is invalid, please check your input");
            }
            int max1 = input[0], max2 = input[1];
         
            Swap(ref max1, ref max2);
            for (int i = 2; i < input.Length; i++)
            {
                if (input[i] > max2)
                {
                    max2 = input[i];
                    Swap(ref max1, ref max2);
                }            
            }
            //if (max1 == max2)
            //{
            //    throw new InvalidOperationException("All the given values are same");
            //}         
            return max2;
        }

        private static void Swap(ref int max1, ref int max2)
        {
            if (max1 < max2)
            {
                int temp = max1;
                max1 = max2;
                max2 = temp;
            }
        }

由这里我们可以类推,如果让你实现寻找第十个大的数,你会如何做呢?