I have a problem with the glfwSetCharCallback function. Whenever I call it, the glfwPollEvents throws an AccessViolationException saying: "Attempted to read or write protected memory. This is often an indication that other memory is corrupt." I have created a very simple wrapper just to demonstrate the problem: using System; using System.Runtime.InteropServices; using System.Securit
我有一个glfwSetCharCallback函数的问题。 无论何时我调用它，glfwPollEvents都会抛出一个AccessViolationException异常：“尝试读取或写入受保护的内存，这通常表示其他内存已损坏。” 我创建了一个非常简单的包装来展示问题： using System; using System.Runtime.InteropServices; using System.Security; namespace Test { public delegate void GlfwCharCallback(GlfwWindow window, Char character); [StructLa
I have two questions: Question 1 Background : I noticed when looking at the implementation of 'AsEnumerable()' method in LINQ from Microsoft, which was: public static IEnumerable<TSource> AsEnumerable<TSource>(this IEnumerable<TSource> source) { return source; } Question 1: I was expecting some kind of casting or something here , but it simply returns the value
我有两个问题： 问题1背景：我注意到当从Microsoft看LINQ中'AsEnumerable（）'方法的实现时，它是： public static IEnumerable<TSource> AsEnumerable<TSource>(this IEnumerable<TSource> source) { return source; } 问题1：我在这里期待某种投射或某种东西，但它只是返回它传递的值。 这个怎么用 ？ 问题2/3背景：我一直在试图理解协变性，逆变和不变。 我认为，我有一种模糊的理解
OK, so I don't sound like an idiot I'm going to state the problem/requirements more explicitly: Needle (pattern) and haystack (text to search) are both C-style null-terminated strings. No length information is provided; if needed, it must be computed. Function should return a pointer to the first match, or NULL if no match is found. Failure cases are not allowed. This means any a
好的，所以我不会听起来像一个白痴，我会更明确地陈述问题/要求： Needle（pattern）和haystack（要搜索的文本）都是C样式的以null结尾的字符串。 没有提供长度信息; 如果需要，它必须被计算。 函数应返回指向第一个匹配的指针，如果找不到匹配项，则返回NULL 。 不允许出现故障。 这意味着任何具有非恒定（或大恒定）存储要求的算法都需要具有回退情况以用于分配失败（并且后备处理中的性能因此导致最坏情况的性能）。
I wrote a netfilter module, in which I had to write an extra thread to clean up a data structure in regular basis. I used schedule() function after each round of clean up was done from the thread. Before using semaphore locks it was working just fine except "General Protection Fault" was occurring. So I used semaphores to lock and unlock the data structure during both insert and dele
我写了一个netfilter模块，在这个模块中我必须编写一个额外的线程来定期清理数据结构。 在每一轮清理完成后，我使用schedule()函数从线程完成。 在使用信号量锁之前，除了"General Protection Fault"发生之外，它工作得很好。 所以我使用信号量在insert和delete操作期间锁定和解锁数据结构。 现在它显示了一个错误，告诉"BUG: Scheduling while atomic" 。 在谷歌搜索后，我得到它显示，因为明确调用sc
Okay, so I'll admit up front this one's going to be a bit long. I'm writing a chess engine for C#, with the eventual goal including UCI implementation. I've got it to the point where, given a board, the engine will generate a list of all valid moves; however, my evaluation code seems to be struggling, because when playing a game against itself, the engine will move two pawns on
好的，所以我会承认，这件事会持续一段时间。 我正在为C＃编写一个国际象棋引擎，其最终目标包括实施UCI。 我已经达到了这样的程度，在给定董事会的情况下，引擎会生成所有有效移动的列表; 然而，我的评估代码似乎很挣扎，因为当与自己对抗时，引擎会移动两边的两个棋子，然后在两边来回移动棋子。 我将概述下面的程序的关键部分，以便最好地让你理解我的代码在什么条件下被调用和使用，希望它能帮助你回答我的问题。 这只
I am currently developing a chess engine in C# and I have hit a bit of a brick wall in developing the code to determine future mobility of any given chess piece in 1, 2 and 3 moves. The basic idea is to reward pieces with a bonus for increased mobility and penalise pieces with less mobility. The chess board is represented as an array of 64 squares, starting from 0 (a8) through 63 (h1), eg Pi
我目前正在用C＃开发一个国际象棋引擎，并且在开发代码以确定任何给定棋子的未来移动性时，我已经碰到了一堵砖墙，在1,2和3步中。 基本的想法是奖励奖品以增加行动力，并惩罚行动不便的人。 国际象棋棋盘被表示为从0（a8）到63（h1）开始的64个正方形的阵列，例如 Piece[] _chessboard = new Piece[64]; 我以这个棋盘位置为例： Black Rooks on squares 3 & 19 (d8 & d6) Black King on square 5 (f8) Black Knight on s
I've seen a few different ways to iterate over a dictionary in C#. Is there a standard way? foreach(KeyValuePair<string, string> entry in myDictionary) { // do something with entry.Value or entry.Key } If you are trying to use a generic Dictionary in C# like you would use an associative array in another language: foreach(var item in myDictionary) { foo(item.Key); bar(item.Va
我已经看到了几种不同的方式来遍历C＃中的字典。 有没有标准的方法？ foreach(KeyValuePair<string, string> entry in myDictionary) { // do something with entry.Value or entry.Key } 如果您正在尝试在C＃中使用通用字典，就像使用其他语言的关联数组一样： foreach(var item in myDictionary) { foo(item.Key); bar(item.Value); } 或者，如果您只需要迭代密钥集合，请使用 foreach(var item in myDicti
This question already has an answer here: Variable definition inside switch statement 5 answers Variable Definition Ignore in C [duplicate] 7 answers Why can't variables be declared in a switch statement? 23 answers In this case , it is not undefined behaviour. For case 0 (when expr == 0 , which is your case), i gets assigned a value before being used (value being read from). OK,
这个问题在这里已经有了答案： switch语句中的变量定义5个答案 变量定义在C中忽略[重复] 7个答案 为什么不能在switch语句中声明变量？ 23个答案 在这种情况下 ，它不是未定义的行为。 对于case 0 （当expr == 0 ，这是你的情况）， i被赋值之前被使用（正在读取的值）。 好的，详细阐述一下这个片段 switch (expr) { int i; case 0: i = 17; default
Possible Duplicate: Why can't variables be declared in a switch statement? How can a variable be used when its definition is bypassed? #include<stdio.h> int main() { int a=1; switch(a) { int b=20; case 1: printf("b is %dn",b); break; default:printf("%dn",b); break; } return 0; } ran on gc
可能重复： 为什么不能在switch语句中声明变量？ 如何在变量被绕过时使用变量？ #include<stdio.h> int main() { int a=1; switch(a) { int b=20; case 1: printf("b is %dn",b); break; default:printf("%dn",b); break; } return 0; } 在gcc 4.6.3上运行，输出不是20.这里发生了什么？ 在switch语句中初始化变量
Is there a way to fall through multiple case statements without stating case value: repeatedly? I know this works: switch (value) { case 1: case 2: case 3: //do some stuff break; case 4: case 5: case 6: //do some different stuff break; default: //default stuff break; } but I'd like to do something like this: switch (value) { ca
有没有办法通过多个案例陈述而没有陈述case value:反复？ 我知道这是有效的： switch (value) { case 1: case 2: case 3: //do some stuff break; case 4: case 5: case 6: //do some different stuff break; default: //default stuff break; } 但我想要做这样的事情： switch (value) { case 1,2,3: //Do Something break; case 4,5,6: