What's the best way to call a generic method when the type parameter isn't known at compile time, but instead is obtained dynamically at runtime? Consider the following sample code - inside the Example() method, what's the most concise way to invoke GenericMethod<T>() using the Type stored in the myType variable? public class Sample { public void Example(string typeName)
当编译时不知道类型参数，而是在运行时动态获得类型参数时，调用泛型方法的最佳方式是什么？ 考虑下面的示例代码 - 在Example()方法中，使用存储在myType变量中的Type调用GenericMethod<T>()的最简洁方法是什么？ public class Sample { public void Example(string typeName) { Type myType = FindType(typeName); // What goes here to call GenericMethod<T>()? GenericMetho
I'm getting a warning from ReSharper about a call to a virtual member from my objects constructor. Why would this be something not to do? When an object written in C# is constructed, what happens is that the initializers run in order from the most derived class to the base class, and then constructors run in order from the base class to the most derived class (see Eric Lippert's blog
我从ReSharper收到关于从我的对象构造函数调用虚拟成员的警告。 为什么这是不该做的事情？ 当用C＃编写的对象被构造时，会发生什么是初始化器从大多数派生类到基类的顺序运行，然后构造函数按从基类到最派生类的顺序运行（请参阅Eric Lippert的博客以了解详细信息至于为什么这是）。 在.NET中，对象也不会在构造时改变类型，但是从最为派生的类型开始，方法表用于最派生的类型。 这意味着虚拟方法调用总是运行在最派生
I must admit, that usually I haven't bothered switching between the Debug and Release configurations in my program, and I have usually opted to go for the Debug configuration, even when the programs are actually deployed at the customers place. As far as I know, the only difference between these configurations if you don't change it manually is that Debug have the DEBUG constant defined
我必须承认，通常我不会在我的程序中调试和发布配置之间切换，而且我通常选择进行调试配置，即使这些程序实际部署在客户的地方。 据我所知，这些配置之间的唯一区别是，如果您不手动更改它，则Debug将定义DEBUG常量，并且Release将检查Optimize代码。 所以我的问题实际上是双重的： 这两种配置有很大的性能差异。 是否有任何特定类型的代码会导致性能差异很大，还是实际上并不那么重要？ 是否有任何类型的代码在Debug配
I have not realized that there is a difference between stack an heap allocation in C. I have written a seriously big program in with stack allocated arrays, but obviously they are not sufficiently big to store the read data. Thus, I need to rewrite everything with malloc allocation. Is there a clever way how 2D arrays can be allocated dynamicaly to heap, and their usage in the code to be simila
我还没有意识到在C中堆栈分配是有区别的。我已经用堆栈分配数组编写了一个严重的大程序，但显然它们不足以存储读取的数据。 因此，我需要用malloc分配重写所有内容。 有没有一种聪明的方式如何将二维数组动态分配到堆中，以及它们在代码中的用法是否相似，以便分配堆栈，这意味着： 我的代码看起来像这样： int MM,NN; float Edge[MM][NN]; Do_Something(MM,NN,Edge); 被调用的过程定义如下： void Do_Something(int MM,in
Is there a language where we can enable a mechanism that allocates new stack space on the heap when the original stack space is exceeded? I remember doing a lab in my university where we fiddled with inline assembly in C to implement a heap-based extensible stack, so I know it should be possible in principle. I understand it may be useful to get a stack overflow error when developing an app b
有没有一种语言，我们可以启用一个机制，当超过原始堆栈空间时，在堆上分配新的堆栈空间？ 我记得在我的大学做了一个实验室，在那里我们用C中的内联汇编来实现一个基于堆的可扩展堆栈，所以我原则上知道它应该是可能的。 我知道在开发应用程序时得到堆栈溢出错误可能很有用，因为它会在不使系统占用大量内存并开始交换的情况下快速终止疯狂的无限递归。 但是，如果您已经完成了经过良好测试的应用程序，并且希望它尽可能健
I have been looking through the SDL & Quake (1/2) source code and I am wondering why the coders find a need to implement their own malloc? (I think the SDL_malloc is just for internal SDL use). And how? Do their functions call the C malloc to allocate a huge block on the heap and then manage their memory there? Or maybe they statically allocate a huge array (but that looses flexibility)?
我一直在浏览SDL和Quake（1/2）源代码，我想知道为什么编程人员需要实现他们自己的malloc？ （我认为SDL_malloc仅供内部SDL使用）。 如何？ 他们的函数是否调用C malloc来在堆上分配一个大块，然后在那里管理它们的内存？ 或者，也许他们静态地分配一个巨大的数组（但是失去了灵活性）？ 这是我最感兴趣的（也适用于普通的malloc）：他们自己是否只管理堆？ 如果是这样，他们怎么知道堆的开始和结束在内存中？ 他们将如
It's known that calloc is different than malloc in that it initializes the memory allocated. With calloc , the memory is set to zero. With malloc , the memory is not cleared. So in everyday work, I regard calloc as malloc + memset . Incidentally, for fun, I wrote the following code for a benchmark. The result is confusing. Code 1: #include<stdio.h> #include<stdlib.h> #de
众所周知， calloc与malloc不同，因为它初始化分配的内存。 使用calloc ，内存设置为零。 使用malloc ，内存不会被清除。 所以在日常工作中，我将calloc视为malloc + memset 。 顺便提一下，为了好玩，我为基准编写了以下代码。 结果令人困惑。 代码1： #include<stdio.h> #include<stdlib.h> #define BLOCK_SIZE 1024*1024*256 int main() { int i=0; char *buf[10]; while(i<
This has been something that has bothered me for ages now. We are all taught in school (at least, I was) that you MUST free every pointer that is allocated. I'm a bit curious, though, about the real cost of not freeing memory. In some obvious cases, like when malloc is called inside a loop or part of a thread execution, it's very important to free so there are no memory leaks. But co
这已经困扰了我很多年了。 我们都在学校教书（至少，我是），你必须释放分配的每一个指针。 但我有点好奇，关于不释放内存的真正代价。 在一些明显的情况下，例如malloc在循环或线程执行的一部分内被调用时，释放非常重要，这样就没有内存泄漏。 但请考虑以下两个示例： 首先，如果我有这样的代码： int main() { char *a = malloc(1024); /* Do some arbitrary stuff with 'a' (no alloc functions) */ retu
I am really new to C, so I am sorry if this is a absolute beginner question, but I am getting a segmentation error when I am building large array, relevant bits of what I am doing is: unsigned long long ust_limit; unsigned long long arr_size; /* ust_limit gets value around here ... */ arr_size = ((ust_limit + 1) / 2) - 1; unsigned long long numbs[(int)arr_size]; This works for some values of
我对C非常陌生，所以如果这是一个绝对的初学者问题，我很抱歉，但是当我构建大型数组时，我遇到了分割错误，我正在做的是相关的部分： unsigned long long ust_limit; unsigned long long arr_size; /* ust_limit gets value around here ... */ arr_size = ((ust_limit + 1) / 2) - 1; unsigned long long numbs[(int)arr_size]; 这适用于ust_limit的某些值，但是当它达到大约4.000.000以上时会出现分段错误。 我想要的是
Suppose you are running a program with interrupts handling enabled on a processor. Instruction Pointer points to zero. How can we get to know the cause that caused the Instruction Pointer to point to 0. I'm not clear whether is it something related to the location of ISRs? As far as I know in some of the processors, IP=0 means the reset address. But why would a running program goto the
假设您正在运行一个处理器上启用了中断处理的程序。 指令指针指向零。 我们如何才能知道导致指令指针指向0的原因。 我不清楚这是否与ISR的位置有关？ 据我所知，在某些处理器中，IP = 0意味着复位地址。 但为什么一个正在运行的程序转到了这个地址？ 什么都可能是导致IP指向0的原因？ 基本上所有的jmp指令和ret都可以跳转到0.例子： jnz 0 ;; encoded as relative jump JNZ -(next IP) jmp 00000000 ;; a