bit

MSBuild and $(ProgramFiles) issue with 32/64 bits

匿名 (未验证) 提交于 2019-12-03 02:51:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I wrote a custom MSBuild task which calls SubWCRev.exe , an executable which (typically) resides in C:\Program Files\TortoiseSVN\bin , whether it's 32 or 64 bits, since TortoiseSVN provides both versions. The problem is, Visual Studio 2010 only has a 32 bit version. So when my colleagues with a 64 bit box try to build using my shiny new task, $(ProgramFiles) resolves to C:\Program Files(x86) and it explodes saying that SubWCRev.exe cannot be found. Because they have the 64 bit version of TortoiseSVN, which lives in C:\Program Files ! Is

cx_Oracle - DLL load failed

匿名 (未验证) 提交于 2019-12-03 02:51:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I have a problem importing cx_Oracle with Python. I know a lot of issues with cx_Oracle have been discussed here, but it seems that I cannot find a solution to my problem after reading all the related topics. I have two machines, one is my computer and another one is a remote workstation, which have similar configs (Windows 7, 64-bits). I need to install cx_Oracle on the remote workstation but it does not work, whereas it works fine on my computer (I can import the module successfully and connect to my DB). On the remote workstation, I have

Rotating a bitmap 90 degrees

匿名 (未验证) 提交于 2019-12-03 02:50:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I have a one 64-bit integer , which I need to rotate 90 degrees in 8 x 8 area (preferably with straight bit-manipulation). I cannot figure out any handy algorithm for that. For instance, this: // 0xD000000000000000 = 1101000000000000000000000000000000000000000000000000000000000000 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 after rotation becomes this: // 0x101000100000000 = 0000000100000001000000000000000100000000000000000000000000000000 0 0 0 0 0 0 0 1 0 0

Content Transfer Encoding 7bit or 8 bit

匿名 (未验证) 提交于 2019-12-03 02:50:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: While sending email content, it is required to set "Content Transfer Encoding" header. I observed many headers of emails that I received. Some emails using "7bit" and some are using "8bit". What is the difference between these two? Which is recommended? Is there any special encoding required for email body in order to set these headers? 回答1: It can be a bit dense to read, but the "Content-Transfer-Encoding" section of RFC 1341 has all of the details: http://www.w3.org/Protocols/rfc1341/5_Content-Transfer-Encoding.html The situation kinda

Radix sort using bitwise operations

匿名 (未验证) 提交于 2019-12-03 02:50:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: First of all this is homework , and I found another topic talking about the same subject but there was no answer. Here is the problem: The main problem is how to make this kind of sort. Should I convert each integer to bits and compare them? Please do not give me the solution just a hint or an explanation of how to do it. Thanks for your help ! [EDIT] I found this script in the internet but i did not understand how it works : #include <cstdlib> #include <iostream> #include <string> #include <cctype> #include<algorithm> #include<string>

Dividing by power of 2 using bit shifting

匿名 (未验证) 提交于 2019-12-03 02:49:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I've got the following task: Compute x/(2^n) , for 0 <= n <= 30 using bit shifting. Requirement: Round toward zero. Examples: divpwr2(15,1) = 7 divpwr2(-33,4) = -2 Legal operators: ! ~ & ^ | + << >> Maximum number of operators: 15 Here is what I've got so far: public int DivideByPowerOf2(int x, int n) { //TODO: find out why DivideByPowerOf2(-33,4) = -3 instead of -2 return x >> n; } DivideByPowerOf2(15,1) = 7 is ok. But DivideByPowerOf2(-33,4) = -3 instead of -2. Why? 回答1: After looking for a good answer myself, I stumbled across this and

How is infinity represented in a C double?

匿名 (未验证) 提交于 2019-12-03 02:49:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I learned from the book Computer Systems: A Programmer's Perspective that the IEEE standard requires the double precision floating number to be represented using the following 64-bit binary format: s: 1 bit for sign exp: 11 bits for exponent frac: 52 bits for fraction The +infinity is represented as a special value with the following pattern: s = 0 all exp bits are 1 all fraction bits are 0 And I think the full 64-bit for double should be in the following order: (s)(exp)(frac) So I write the following C code to verify it: //Check the

Microsoft.ACE.OLEDB.12.0 is not registered

匿名 (未验证) 提交于 2019-12-03 02:49:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I have a SQL Server job that runs monthly that runs in server. Job is using an SSIS package and is supposed to extract the data from database and and create an Excel sheet and copy the data into Excel 2003. I actually got around 140,000 rows from the database due to truncation issue in Excel 2003 (Excel supports 64,000 rows). So I modified the config file to support 2007 Excel format. "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" + @[User::FullPath] + ";Extended Properties=\"Excel 12.0;HDR=YES\" But when I try to execute the job, it fails

Where can I find a bit shifting guide for C? [closed]

匿名 (未验证) 提交于 2019-12-03 02:49:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I have looked at What are bitwise shift (bit-shift) operators and how do they work? but I still find the concept of bit shifting difficult to understand. Can someone point me in the direction of a more basic guide to bit shifting in C. I expect it will be something really long since it will need to cover the whole subject. I really do not understand it but I want to learn it, so any help would be appreciated. I am learning the k&r, and that is what this is for so I can do the exercises. I understand the basics, but I still cannot do correct

How do I count the number of zero bits in an integer?

匿名 (未验证) 提交于 2019-12-03 02:48:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: How would i go about finding the number of 'zero' bits in C++. Suppose I have an integer; int value = 276; For which I have the bits 100010100, but how do I count the zeros? 回答1: The easiest most naive way is to just iterate over the bits and count: size_t num_zeroes = 0; for(size_t i = 0; i < CHAR_BIT * sizeof value; ++i) { if ((value & (1 << i)) == 0) ++num_zeroes; } There are all number of better (for different values of "better") ways, but this is quite clear, very terse (code-wise), and doesn't require a bunch of setup. One micro