I'm going to be trying to give a talk on async - await and I'm creating a flow chart that attempts to show the possible orders of execution. I attempted to base that off the paragraphs The beginning of an async method is executed just like any other method. That is, it runs synchronously until it hits an “await” (or throws an exception). The “await” keyword is where things can get
我将尝试就async await进行讨论，并创建一个流程图，试图显示可能的执行顺序。 我试图从段落中找出它 异步方法的开始与其他方法一样执行。 也就是说，它同步运行，直到它遇到“等待”（或抛出异常）。 “await”关键字是事物可以异步的地方。 等待就像一个一元运算符：它只需要一个参数，一个等待（一个“awaitable”是一个异步操作）。 等待检查，看看它是否已经完成; 如果awaitable已经完成，那么方法会继续运行（同步，就
I'm trying to get my head around this code: [TestFixture] public class ExampleTest { [Test] public void Example() { AwaitEmptyTask().Wait(); } public async Task AwaitEmptyTask() { await new Task(() => { }); } } The method Example never ends and blocks forever. Why?? The fix (from Stubbing Task returning method in async unit test) is to
我试图让我的头绕过这段代码： [TestFixture] public class ExampleTest { [Test] public void Example() { AwaitEmptyTask().Wait(); } public async Task AwaitEmptyTask() { await new Task(() => { }); } } 方法Example永远不会结束并永远阻塞。 为什么？？ 修复 （从异步单元测试中的Stubbing Task返回方法）是用return Task.FromResult<object>(null);替
I have the following code and just want to make sure I have the concept of multithreading down on a high level. public async Task<List<Category>> GetProjectsByCategoryIDAsync(Int16 categoryid) { try { using (YeagerTechEntities DbContext = new YeagerTechEntities()) { DbContext.Configuration.ProxyCreationEnabled = false;
我有下面的代码，只是想确保我有一个高层次的多线程的概念。 public async Task<List<Category>> GetProjectsByCategoryIDAsync(Int16 categoryid) { try { using (YeagerTechEntities DbContext = new YeagerTechEntities()) { DbContext.Configuration.ProxyCreationEnabled = false; DbContext.Database.Connection.Open();
the new async/await keywords in C# 5 look very promising but I read an article about the performance impact on those applications since the compiler will generate a quite complex state machine for async methods. Async-programming using these keywords is so much easier but is it as good as say SocketAsyncEventArgs for Sockets ? Second question: Are asynchronous IO methods like Stream.WriteAsyn
C＃5中新的异步/等待关键字看起来非常有前途，但我阅读了一篇关于对这些应用程序的性能影响的文章，因为编译器将为异步方法生成一个相当复杂的状态机。 使用这些关键字的异步编程非常容易，但是与套接字的SocketAsyncEventArgs一样好？ 第二个问题：像Stream.WriteAsync这样的异步IO方法真的是异步的（在.Net上的完成端口或Mono上的epoll / poll），还是这些方法将写入调用推送到线程池是便宜的包装？ 第三个问题：在UI应
Browsing through the channel 9 msdn videos I found the following unanswered comment and was hoping someone could possibly explain it? I dont get the point of the async keyword. Why not just allow the await keyword anytime the method returns Task, just like iterators can yield return on any method that returns an IEnumerable. I'm sure there is a good reason, I'd just like to understan
通过频道浏览9个msdn视频我发现了以下未答复的评论，并希望有人可以解释它？ 我没有得到异步关键字的重点。 为什么不在方法返回Task时只允许await关键字，就像迭代器可以返回任何返回IEnumerable的方法的返回值一样。 我确信有一个很好的理由，我只想明白为什么上述建议不可行。 它主要是为了避免向后兼容性问题而引入的。 如果编译器必须推断方法的async （即通过检测await关键字），那么存在一些细微的场景，其中现有
I have the following tables in my database, all with the same table name but different schema's. dbo.Versions bpm.Versions wf.Version ... All x.Versions have a FK to the Version table. I've created the POCO classes from it (this gave me classes like Version, Version1, .... - - I've renamed the classNames to Version and BPMVersion, .... but the mapping still exists to the r
我在我的数据库中有以下表格，所有表格都有相同的表格名称，但有不同的模式。 dbo.Versions bpm.Versions wf.Version ... 所有x.Versions都有一个FK到版本表。 我已经创建了POCO类（这给了我像Version，Version1，...的类 - - 我已经将classNames重命名为Version和BPMVersion，但映射仍然存在于右表中。 这是我映射到bpm.Versions的BPMVersion映射的一个示例 // Primary Key this.HasKey(t => t.Id); // Pro
I am developing a originally serial code in KSVD package to support OpenMP. The original code, which serves like im2col in MATLAB and extracts patches from the image, is shown as follows: /* n stands for the size of an image, sz stands for the patch size to extract */ int blocknum = 0; for (k=0; k<=n[2]-sz[2]; k+=1) { for (j=0; j<=n[1]-sz[1]; j+=1) { for (i=0; i<=n[0]-sz[0]
我正在KSVD软件包中开发一个原本的串行代码来支持OpenMP。 原始代码在MATLAB中用作im2col并从图像中提取补丁，如下所示： /* n stands for the size of an image, sz stands for the patch size to extract */ int blocknum = 0; for (k=0; k<=n[2]-sz[2]; k+=1) { for (j=0; j<=n[1]-sz[1]; j+=1) { for (i=0; i<=n[0]-sz[0]; i+=1) { /* copy single block */ for (m=0; m<
I recently started working with Intel MPI for parallelizing a very simple flow solver. I'm wondering whether I should use the stack for storing my variables (ie use datatype name[size]; for declaration) or the heap (ie use datatype *name = (datatype *)malloc(size*sizeof(datatype)); ). First I used malloc, because I split up the flow field into n parts, where n is the number of processes th
我最近开始与Intel MPI合作，对一个非常简单的流程求解器进行并行化。 我想知道我是否应该使用栈来存储我的变量（即使用datatype name[size];对于声明）或堆（即使用datatype *name = (datatype *)malloc(size*sizeof(datatype)); ）。 首先我使用了malloc，因为我将流场分成了n部分，其中n是创建的进程数，我认为对于n的所有值使用相同的代码会很好。 这意味着我的数组的大小在运行时首先已知。 所以我显然需要动态内存分
I'm currently learning C and I'm confused about memory layout and pointers. In the following code, it is my understanding that the array is allocated on the stack. #include <stdio.h> int main () { int x[4]; x[0] = 3; x[1] = 2; x[2] = 1; printf("%pn",x); printf("%pn", &x); } My question is, why do the two print calls output the same value? I tried a similar sni
我目前正在学习C语言，并对内存布局和指针感到困惑。 在下面的代码中，我的理解是数组在堆栈上分配。 #include <stdio.h> int main () { int x[4]; x[0] = 3; x[1] = 2; x[2] = 1; printf("%pn",x); printf("%pn", &x); } 我的问题是，为什么这两个打印调用输出相同的值？ 我尝试了一个类似的片段使用malloc（分配在堆上），并且值不同。 #include <stdio.h> #include <stdlib.h> int
I try to understand something about the amount of memory allocated to the stack and to the heap. Suppose sizeof(char) = 1 byte, and sizeof(void *) = 4 byte. given the following code: #include<stdio.h> #include<stdlib.h> #include<string.h> int main() { int i; char *str1 = "hello"; char str2[] = "hello"; char *str3 = (char*)malloc(strlen(str2)); //free code
我尝试了解分配给堆栈和堆的内存量。 假设sizeof（char）= 1个字节，sizeof（void *）= 4个字节。 给出以下代码： #include<stdio.h> #include<stdlib.h> #include<string.h> int main() { int i; char *str1 = "hello"; char str2[] = "hello"; char *str3 = (char*)malloc(strlen(str2)); //free code return 0; } 我们被告知分配给堆的内存量是5个字节，我明白这确实是malloc