Nested TransactionScope for NUnit TestFixure and SetUp

I derive from this base class in order to enclose each indivdual test into a transaction that is rolled back public abstract class TransactionBackedTest { private TransactionScope _transactionScope; [SetUp] public void TransactionSetUp() { var transactionOptions = new TransactionOptions { IsolationLevel = IsolationLevel.ReadCommitted, Time

用于NUnit TestFixure和SetUp的嵌套TransactionScope

我从这个基类中派生出来,以便将每个单独的测试放入一个回退的事务中 public abstract class TransactionBackedTest { private TransactionScope _transactionScope; [SetUp] public void TransactionSetUp() { var transactionOptions = new TransactionOptions { IsolationLevel = IsolationLevel.ReadCommitted, Timeout = TransactionManager.MaximumTimeout

C default arguments

有没有一种方法可以在C中为函数指定默认参数? Not really. The only way would be to write a varargs function and manually fill in default values for arguments which the caller doesn't pass. Wow, everybody is such a pessimist around here. The answer is yes. It ain't trivial: by the end, we'll have the core function, a supporting struct, a wrapper function, and a macro around the wra

C默认参数

有没有一种方法可以在C中为函数指定默认参数? 不是真的。 唯一的办法是编写一个可变参数函数并手动为调用者不传递的参数填充默认值。 哇,这里每个人都是这样的悲观主义者。 答案是肯定的。 它不是微不足道的:最终,我们将拥有核心函数,支持结构,包装函数和围绕包装函数的宏。 在我的工作中,我有一组宏来自动化所有这些; 一旦你了解了流程,你也很容易做到这一点。 我在其他地方写过这篇文章,所以这里有一个详细

Understanding and implementing malloc

How is malloc implemented internally ? How to implement malloc with below necessary conditions • Malloc allocates at least the number of bytes requested • The pointer returned by malloc points to an allocated space ( ie a space where the program can read or write successfully;) • No other call to malloc will allocate this space or any portion of it, unless the pointer has been freed before

理解和实现malloc

malloc是如何在内部实现的? 如何在以下必要条件下实现malloc •Malloc至少分配请求的字节数 •malloc返回的指针指向已分配的空间(即程序可以成功读取或写入的空间;) •除非指针已被释放,否则不会有其他任何对malloc的调用分配此空间或其任何部分。 •malloc应该是可处理的:malloc必须尽快终止(它不应该是NP-hard!;) •Malloc还应提供调整和释放。 该函数服从以下签名:void * malloc(size_t size); brk()

what does casting a pointer 'actually' do under the hood?

Say I have the following c code: int* vector = (int*)malloc(5 * sizeof(int)); malloc returns a void pointer because doesn't know what is being asked to allocate space for. Therefore, we are casting the void pointer to an int pointer. Does the cast actually do anything at runtime, or is it just required for compiling? If it does work at runtime, what is it doing? Casting of pointers i

什么确实在铸造一个指针?

假设我有以下c代码: int* vector = (int*)malloc(5 * sizeof(int)); malloc返回一个无效指针,因为不知道什么被要求分配空间。 因此,我们将void指针投射到一个int指针。 转换实际上是否在运行时执行任何操作,还是仅仅需要编译? 如果它在运行时工作,它在做什么? 指针的转换在编译时是必需的,除了指向void的指针之外,可能会将指针转换为指针或将指针转换为任何类型,而无需显式强制转换。 运行时发生的情况不是

Function Pointer cast at declaration

While looking for informations about calloc, I founded in the source code: char *malloc(); in the calloc function. Does it cast the void *malloc(size_t) into a function pointer returning a char* ? This syntaxe does not compile for me. In the old days, before ANSI C provides void * as the generic pointer type, char * was used for this purpose. The code is from the source of Version 7 Uni

函数指针在声明时强制转换

在寻找有关calloc的信息时,我创建了源代码: char *malloc(); 在calloc函数中。 它是否将void * malloc(size_t)转换为返回char *的函数指针? 这个语法不适合我编译。 在过去,在ANSI C提供void *作为通用指针类型之前, char *被用于此目的。 代码来自版本7 Unix的源代码,它于1979年发布(在ANSI C之前)。 这就是为什么需要在旧的ANSI代码中投射malloc()的返回值。 参考:C FAQ 这来自一些非常古老的源代

there is no heap in c?

I just started reading The C Programming Language by Brian Kernighan and Dennis Ritchie, and I found this statement: The language does not define any storage allocation facility other than static definition and the stack discipline provided by the local variables of functions; there is no heap or garbage collection. So does this mean that it is due to the malloc() function, which returns som

c中没有堆?

我刚开始阅读Brian Kernighan和Dennis Ritchie的C编程语言,并且我发现了这样一段话: 除了静态定义和由函数的局部变量提供的堆栈规则以外,该语言没有定义任何存储分配工具; 没有堆或垃圾收集。 那么这是否意味着它是由malloc()函数产生的,该函数从堆中返回一些内存地址,C可以访问堆内存? 然后必须用其他语言编写malloc ,最可能是程序集或B? 这可能是一个愚蠢的疑问,但我必须清除它。 谢谢。 C语言本身并不直

Need assistance in understanding this code using malloc and pointers

Here is a little snippet of code from Wikipedia's article on malloc(): int *ptr; ptr = malloc(10 * sizeof (*ptr)); // Without a cast ptr = (int*)malloc(10 * sizeof (int)); // With a cast I was wondering if someone could help me understand what is going on here. So, from what I know, it seems like this is what's happening: 1) initialize an integer pointer that points to NU

需要使用malloc和指针来理解这段代码

以下是维基百科关于malloc()的文章中的一小段代码: int *ptr; ptr = malloc(10 * sizeof (*ptr)); // Without a cast ptr = (int*)malloc(10 * sizeof (int)); // With a cast 我想知道是否有人能帮助我理解这里发生了什么。 所以,从我所知道的情况来看,似乎是这样发生的: 1)初始化一个指向NULL的整数指针。 它是一个指针,所以它的大小是4字节。 解引用这个指针将返回值NULL。 2)由于C允许这种类

C language: pointer returned by malloc() and casting

Possible Duplicate: Do I cast the result of malloc? What is the Best Practice for malloc? Im new with c language and i have a question as i was explained in class: The type of the pointer returned by malloc() is void *, and therefore we need to cast it into the required type which means i need to do: char *str; str = (char *) malloc(14); but i dont understand, is it a must? lets say

C语言:由malloc()返回的指针和转换

可能重复: 我输入malloc的结果吗? malloc的最佳实践是什么? 即时通讯新的与C语言,我有一个问题,因为我在课堂上解释: malloc()返回的指针的类型是void *,因此我们需要将它转换为所需的类型 这意味着我需要这样做: char *str; str = (char *) malloc(14); 但我不明白,这是必须的吗? 让我们说malloc返回的是void指针,为什么我看到很多没有做铸造的例子呢? 一个例子: float *arr = malloc (20 * sizeof

Understanding malloc

I was given this piece of example code on a lab and I'm trying to understand it int size = 5; int **ppi2 = (int **) malloc(size * sizeof(int *)); Here's my breakdown of it as of now: In order to allocate sufficient memory, we must multiply the number of things we want to allocate memory for (size) by the size of those things (sizeof(int *)). That much is straightforward and makes a l

了解malloc

我在实验室获得了这段示例代码,并试图理解它 int size = 5; int **ppi2 = (int **) malloc(size * sizeof(int *)); 以下是截至目前的情况: 为了分配足够的内存,我们必须将我们想要为(size)分配内存的东西的数量乘以这些东西的大小(sizeof(int *))。 这很简单,很有意义。 我知道我们必须将malloc的结果转换为双指针,因为malloc返回字节并且变量是双指针,但为什么我们首先将ppi2设置为等于malloc的结果? 它是

Casting as Struct Pointer + Malloc

I have some ideas of what this would do, but I'd like a more professional and experienced explaination. typedef struct{ char str[50]; unsigned short num; }s; s *name = (s *) malloc(sizeof(s)); The first part defines a struct. I have no problem with that This will create a pointer to "s" struct. malloc() returns a memory address that will be cast as a pointer to &quo

施放为结构指针+ Malloc

我对这会做什么有一些想法,但我想要更专业和有经验的解释。 typedef struct{ char str[50]; unsigned short num; }s; s *name = (s *) malloc(sizeof(s)); 第一部分定义了一个结构。 对于那件事我没有任何疑问 这将创建一个指向“s”结构的指针。 malloc()返回一个内存地址,该地址将被转换为指向“s”的指针。 然而, sizeof(s)我相信我有一些问题的理解。 char str[50]; = 50个字节。 unsigned short num;