C# bitwise shift on ushort (UInt16)

I need to perform a bitwise left shift on a 16-bit integer (ushort / UInt16), but the bitwise operators in C# seem to apply to int (32-bit) only. How can I use << on an ushort, or at least get to the same result with a simple workaround? 移位后将结果值转换回ushort: ushort value = 1; ushort shifted = (ushort)(value << 2);

ushort上的C#按位移动(UInt16)

我需要对16位整数(ushort / UInt16)执行按位左移,但C#中的按位运算符似乎仅适用于int(32位)。 我怎样才能在一个ushort上使用<<,或者至少通过一个简单的解决方法获得相同的结果? 移位后将结果值转换回ushort: ushort value = 1; ushort shifted = (ushort)(value << 2);

Using bitwise operators

I've been studying C# and ran accross some familiar ground from my old work in C++. I never understood the reason for bitwise operators in a real application. I've never used them and have never had in a reason to use them. I've been studying how they work; the example below shows the shift bitwise operator. What is the point of bitwise operators, their use and how they work? M

使用按位运算符

我一直在研究C#,并在C ++的旧作中熟悉了一些熟悉的内容。 我从来没有理解在实际应用中按位运算符的原因。 我从来没有使用过它们,从来没有理由使用它们。 我一直在研究他们的工作方式。 下面的例子显示了移位位运算符。 按位运算符,它们的用途以及它们的工作原理是什么? 也许我在按位逻辑中错过了一些东西。 byte bitComp = 15; // bitComp = 15 = 00001111b byte bresult = (byte) ~bitComp; // bresul

define SOMETHING (1 << 0)

I came accros this line of code: #define CPARSER_FLAGS_DEBUG (1 << 0) What does it do? Its the same as: #define CPARSER_FLAGS_DEBUG (1) Right? In C-inspired languages, << and >> operators are left and right bitwise shift operators (though in C++ they can be overloaded — the most famous overload is probably I/O stream operators). For example, x = y <<

定义SOMETHING(1 << 0)

我来accros这行代码: #define CPARSER_FLAGS_DEBUG (1 << 0) 它有什么作用? 它与以下相同: #define CPARSER_FLAGS_DEBUG (1) 对? 在C启发式语言中, <<和>>运算符是左右位运算符(尽管在C ++中它们可以被重载 - 最着名的重载可能是I / O流运算符)。 例如, x = y << 2; 指定x将y向左移两位的结果。 通常你会在低级代码中看到很多字节移位,这就是为什么......任何硬件都

Ninject WCF bootstrapper registering multiple services

I'm having a problem getting the wcf extensions to work with more than one self host bootstrapper. With one my services are created by ninject fine (per call), but when I add another I get an exception that the ChannelDispatcher is unable to open its IChannelListener, the inner exception states that a registration already eixsts for URI 'net.tcp://localhost:901/MyService'. My regis

Ninject WCF引导程序注册多个服务

我有一个问题让wcf扩展与多个自我主机引导程序一起工作。 有一个我的服务是由ninject罚款(每次调用)创建的,但是当我添加另一个我得到一个异常,ChannelDispatcher无法打开它的IChannelListener时,内部异常指出注册已经为URI'net.tcp:/ /本地主机:901 /为MyService”。 我的注册码如下所示: var myService= NinjectWcfConfiguration.Create<MyService, NinjectServiceSelfHostFactory>(); _myServiceHost= new

XML encryption and decryption for multiple recipients with X509 certificates

I have managed to encrypt and decrypt xml documents using examples on MSDN. http://msdn.microsoft.com/en-us/library/ms229744.aspx and http://msdn.microsoft.com/en-us/library/ms229943.aspx This is all done according to the W3C XML encryption standard(XML Enc). It all works good. My problem is that one xml document is intended for 2 or 3 recipients. I want to encrypt same xml with multiple k

使用X509证书对多个收件人进行XML加密和解密

我已经设法使用MSDN上的示例加密和解密xml文档。 http://msdn.microsoft.com/en-us/library/ms229744.aspx和http://msdn.microsoft.com/en-us/library/ms229943.aspx 这一切都是根据W3C XML加密标准(XML Enc)完成的。 这一切都很好。 我的问题是,一个XML文档是为2或3收件人。 我想用多个密钥(X509证书公钥)加密相同的XML,以便文档可以被多个收件人解密。 根据W3C XML加密标准,通过使用包含加密的对称会话密钥的

Regular expression for parsing similar assembler instructions

The intro is a bit lengthy, so please bear with me. :) I am writing a simple regex-based parser for a large source file written in assembler. Most of these instructions are just moving, adding, subtracting and jumping around, but it's a pretty large file which I need to port to two different languages and I am too lazy to do it manually. That's the requirement and I can't do much

正则表达式用于解析相似的汇编程序指令

介绍有点冗长,请耐心等待。 :) 我正在为使用汇编编写的大型源文件编写一个简单的基于正则表达式的解析器。 大多数这些指令只是移动,增加,减少和跳跃,但它是一个非常大的文件,我需要移植到两种不同的语言,我懒得手动完成它。 这是要求,我不能做很多'回合(所以请不要回答“你为什么不简单使用ANTLR”)。 因此,在我做了一些预处理之后(我已经完成了这部分工作:替换了定义和宏,并且删除了多余的空白和注释),

A simple x86 disassembler open source for kernel use

I'm writing a kernel for educational purposes and I want to integrate a disassembler into my kernel. Since I'm going to integrate it into the kernel I want it to be very small and simple, ie I only need it to receive a memory address and return the assembly instruction at that address. I've looked for an open source that can do that but all I found is big libraries which are way

一个简单的x86反汇编程序开源内核使用

我正在编写一个用于教育目的的内核,我想将反汇编程序集成到我的内核中。 由于我要将它集成到内核中,我希望它非常小巧和简单, 即我只需要它接收一个内存地址并返回该地址的汇编指令。 我找过一个可以做到这一点的开源软件,但是我发现所有的大型软件库都太大而且复杂。 我正在谈论几个(2-4)源代码文件,它们不需要复杂的安装/集成,而且它的功能非常有限,只能提供非常基本的功能。 有人知道这样的开源(当然是C)

Using LEA on values that aren't addresses / pointers?

I was trying to understand how Address Computation Instruction works, especially with leaq command. Then I get confused when I see examples using leaq to do arithmetic computation. For example, the following c code, long m12(long x) { return x*12; } In assembly, leaq (%rdi, %rdi, 2), %rax salq $2, $rax If my understanding is right, leaq should move whatever address (%rdi, %rdi, 2), which sh

在不是地址/指针的值上使用LEA?

我试图了解地址计算指令的工作原理,尤其是使用leaq命令。 然后当我看到使用leaq进行算术计算的例子时,我感到困惑。 例如,下面的c代码, long m12(long x) { return x*12; } 在组装中, leaq (%rdi, %rdi, 2), %rax salq $2, $rax 如果我的理解是正确的,leaq应该将任何地址(%rdi,%rdi,2)移到%rax中,这应该是2 *%rdi +%rdi。 我感到困惑的是因为值x存储在%rdi中,这就是存储器地址,为什么%rdi乘以3然后左

Assembly Register Arithmetic?

This question already has an answer here: What's the purpose of the LEA instruction? 14 answers lea does not move addresses, it computes the effective address and stores the resulting address into the destination register. Address computations are performed in machine language with regular arithmetic, not C pointer arithmetic. lea eax,[ebp-40] subtracts 40 from the value in register e

汇编寄存器算术?

这个问题在这里已经有了答案: LEA指令的目的是什么? 14个答案 lea不会移动地址,它会计算有效地址并将结果地址存储到目标寄存器中。 地址计算是用机器语言以常规算术进行的,而不是C指针算术。 lea eax,[ebp-40] 从寄存器ebp的值中减去40 ,并将结果存储在寄存器eax 。 mov DWORD PTR [esp+4], 0x80484c 通过将寄存器esp包含的值加上4来计算目标地址,并将值0x80484c (整数8407116存储为该地址处占用4个字节的32位

How to access locals through stack trace? (Mimicking dynamic scope)

Background Even though it's possible to compile C# code at runtime, it's impossible to include and run the generated code in the current scope. Instead all variables have to be passed as explicit parameters. Compared with dynamic programming languages like Python, one could never truly replicate the complete behaviour of eval (as in this example). x = 42 print(eval("x + 1")) # Print

如何通过堆栈跟踪访问当地人? (模仿动态范围)

背景 尽管可以在运行时编译C#代码,但不可能在当前作用域中包含和运行生成的代码。 相反,所有变量都必须作为显式参数传递。 与Python等动态编程语言相比,人们永远无法真正复制eval的完整行为(如本例中)。 x = 42 print(eval("x + 1")) # Prints 43 这个问题 所以我的问题是(不管它是否真的有用;))是否可以通过使用反射来模仿.NET中的动态范围 。 由于.NET为我们提供了Diagnostics.StackTrace类,它允许我们检