I want to check if an icon exists in the systray; as in, if "X" application has displayed their systray icon in the systray area. I've Googled for information about how to do this but I didn't find anything. UPDATE : This what I've tried in VB.NET translating the C# examples of the url gived by Robert comment, but I don't know how to continue it. Imports System.R
我想检查系统托盘中是否存在图标; 如果“X”应用程序在系统托盘区域中显示了其系统托盘图标。 我已经谷歌了解如何做到这一点的信息，但我没有找到任何东西。 更新： 这是我在VB.NET中尝试翻译Robert评论给出的URL的C＃示例，但我不知道如何继续。 Imports System.Runtime.InteropServices Public Class Form1 Public Declare Function FindWindow Lib "user32.dll" Alias "FindWindowA" (ByVal lpClassName As String
I have a really odd situation using Visual Studio unit testing framework. A test decorated as [TestMethod, ExpectedException(typeof(InvalidOperationException))] fails with System.InvalidOperationException . If I remove all code from the test (to make sure it doesn't throw anything at all) - it passes. It is as if ExpectedException is not even there... If I start an empty project with a
使用Visual Studio单元测试框架的情况非常奇怪。 装饰为[TestMethod, ExpectedException(typeof(InvalidOperationException))] 失败，并显示System.InvalidOperationException 。 如果我从测试中删除所有代码（以确保它不会抛出任何东西） - 它会通过。 就好像ExpectedException甚至不存在... 如果我用一个虚拟测试开始一个空项目，它InvalidOperationException抛出InvalidOperationException它完全按预期工作。 验证Exp
Does anyone have a reference to a poster/one-page pdf or something similar with a list of the eight phases of translation for the C language (the first one being trigraph translation)? I want to have one printed hanging on my wall next to my pc. Update: Sorry for forgetting to specify. I am interested in C90 (although C99 probably is pretty close, _Pragma as mentioned in pmg's answer is C
有没有人可以参考海报/单页pdf或类似C语言翻译八个阶段的列表（第一个是三字母翻译）？ 我想在我的电脑旁边挂一个印在墙上的挂件。 更新：遗忘忘记指定。 我对C90感兴趣（虽然C99可能非常接近， _Pragma Pmg的答案中提到的_Pragma特定于C99，我想避免这种情况）。 获胜的ASCII艺术： ANSI C translation phases ========================= +---------------------
#include <stdio.h> int main(void) { int i = 0; i = i++ + ++i; printf("%dn", i); // 3 i = 1; i = (i++); printf("%dn", i); // 2 Should be 1, no ? volatile int u = 0; u = u++ + ++u; printf("%dn", u); // 1 u = 1; u = (u++); printf("%dn", u); // 2 Should also be one, no ? register int v = 0; v = v++ + ++v; printf("%dn", v); // 3 (Should be the sam
#include <stdio.h> int main(void) { int i = 0; i = i++ + ++i; printf("%dn", i); // 3 i = 1; i = (i++); printf("%dn", i); // 2 Should be 1, no ? volatile int u = 0; u = u++ + ++u; printf("%dn", u); // 1 u = 1; u = (u++); printf("%dn", u); // 2 Should also be one, no ? register int v = 0; v = v++ + ++v; printf("%dn", v); // 3 (Should be the sam
I'm working on a prototype program, which should test different (SessionState, Profile, etc) providers for ASP.NET, ie MySQL, Oracle, etc. At the moment I'm using the MySQL provider. I've just managed to instantiate the provider, a SessionStateContainer and the SessionState itself. Type mySqlType = Type.GetType("MySql.Web.SessionState.MySqlSessionStateStore, MySql.Web, Version=6.6.4
我正在开发一个原型程序，它应该为ASP.NET测试不同的（SessionState，Profile等）提供程序，即MySQL，Oracle等。目前我正在使用MySQL提供程序。 我刚刚设法实例化提供程序，SessionStateContainer和SessionState本身。 Type mySqlType = Type.GetType("MySql.Web.SessionState.MySqlSessionStateStore, MySql.Web, Version=6.6.4.0, Culture=neutral, PublicKeyToken=c5687fc88969c44d", true, true); SessionStateStoreProvid
I saw an interesting technique used in an answer to another question, and would like to understand it a little better. We're given an unsigned 64-bit integer, and we are interested in the following bits: 1.......2.......3.......4.......5.......6.......7.......8....... Specifically, we'd like to move them to the top eight positions, like so: 12345678...................................
我看到一个有趣的技术用于解答另一个问题，并希望更好地理解它。 我们给出了一个无符号的64位整数，我们对以下几点感兴趣： 1.......2.......3.......4.......5.......6.......7.......8....... 具体来说，我们希望将它们移动到前八位，如下所示： 12345678........................................................ 我们不关心由表示的位的值. ，而且他们不必保存。 解决方案是屏蔽不需要的位，并将结果乘以0x2040810
I'm looking for the fastest way to popcount on large buffer of 512 or more bytes. I can guarantee any required alignment, and the buffer size is always a power of 2. The buffer corresponds to block allocations, so typically the bits are either all set, none set, or mostly set favoring the "left" of the buffer, with occasional holes. Some solutions I've considered are: GCC&#
我正在寻找在512或更多字节的大缓冲区上弹出最快的方式。 我可以保证任何需要的对齐，并且缓冲区大小始终是2的幂。缓冲区对应于块分配，因此通常这些比特是全部设置的，没有设置或者主要设置为偏向缓冲区的“左边”，偶尔出现漏洞。 我考虑的一些解决方案是： GCC的__builtin_popcount Bitslice popcount_24words 计数比特集，布赖恩Kernighan的方式 我对最快的解决方案感兴趣，它必须在属于core2或更新版本的32位x86芯
This question already has an answer here: What are bitwise shift (bit-shift) operators and how do they work? 8 answers This is also called the shift operator. It will shift de bits n positions to the left. Shifting the bits by one positions is the same as multiplying or dividing by 2. It is a left-shift operator. It applies on the bit representation of 1 and shift it's bits to left
这个问题在这里已经有了答案： 什么是位移（位移）操作符，它们是如何工作的？ 8个答案 这也被称为移位操作符。 它会将位n位移到左边。 将位移动一个位置与乘以或除以2相同。 它是一个左移操作符。 它适用于1的位表示并将它的位左移2位。 <<是一个二元左移运算符。 左操作数值左移了右操作数指定的位数。 和0填充在右边。 因此， 1即1的二进制移位到左边2个位置，变成100 ，在十进制数中是4 。 所以i被分
This question already has an answer here: What are bitwise shift (bit-shift) operators and how do they work? 8 answers Pay attention to operator precedence. 2 + 2 << 2 is not 2 + (2 << 2) . its same as below :- x + 2 = 4; 4 << 2 ////it means 16 if you perform bitwise on this. Desciption to understand Let shift operation :- Shifts bits to the left. The number to the
这个问题在这里已经有了答案： 什么是位移（位移）操作符，它们是如何工作的？ 8个答案 注意运营商的优先级。 2 + 2 << 2不是2 + (2 << 2) 。 其相同如下： - x + 2 = 4; 4 << 2 ////it means 16 if you perform bitwise on this. 说明理解让移位操作： - 向左移位。 操作员左侧的编号将该编号指定的位数移至右侧。 每次向左移动数字，因此每个左移将原始数字乘以2.使用左移快速乘法或将一组
This question already has an answer here: What are bitwise shift (bit-shift) operators and how do they work? 8 answers Shifting a number to the left is equivalent to multiplying that number by 2n where n is the distance you shifted that number. To see how is that true lets take an example, suppose we have the number 5 so we shift it 2 places to the left, 5 in binary is 00000101 ie 0×27 +
这个问题在这里已经有了答案： 什么是位移（位移）操作符，它们是如何工作的？ 8个答案 向左移动一个数字相当于将该数字乘以2n，其中n是您移动该数字的距离。 为了看看真实情况如何让我们举一个例子，假设我们有数字5所以我们将它向左移2位，二进制中的5是00000101即 0×27 + 0×26 + 0×25 + 0×24 + 0×23 + 1×22 + 0×21 + 1×20 = 1×22 + 1×20 = 4 + 1 = 5 现在， 5 << 2将是00010100即 0×27 + 0×26 + 0×25 + 1×