Devenv.com is the visual studio command line interface such that when you type devenv /? help for devenv appears on the console. However, with no options, devenv.com simply calls devenv.exe (the visual studio GUI).
How do I make a .com program in visual studio 2010?
Thanks
A.
Devenv.com isn't actually an executable that uses the COM file format. It is simply an plain EXE that was just renamed. Windows isn't otherwise fooled by it, it always looks a the header of the file. The "MZ" prefix that's present in the header is enough to convince it that it is really an EXE file and is relocatable, it doesn't just go by the filename extension.
The only reason it exists is as a trick. Based on the way the command processor looks for executable files. If it has a choice between foo.com and foo.exe then it will pick foo.com first.
So when you type "devenv yadayada" at the command prompt then you get the console mode app, devenv.com, which reuses the console window. Which, if necessary, launches devenv.exe
So simply build a normal .exe file. And rename it to .com
Related
My goal is to override the OutputPath property on all projects in a solution to be $(SolutionDir)$(Configuration)\. I would like to set this from within the VS2015 IDE.
I don't want to change the OutputPath in the .csproj or .vcxproj files (I know how to do this and it's not my intention to make a permanent change to the project files). I just need a local change for the moment to build.
I know I can accomplish this from the command line in a Developer Command Prompt by setting /p:OutputPath=$(SolutionDir)$(Configuration)\ on msbuild. Ideally, would like to be able to do this from within the IDE. Is that possible?
I was able to set OutputPath in a Developer Command Prompt and then launch VS2015, open the solution, and build:
"C:\Program Files (x86)\Microsoft Visual Studio 14.0\Common7\IDE\devenv.exe"
Typical way of doing this is to import the same msbuild file in each project, and set OutputPath in that imported file. Disadvantage: project files need to be modified, it's not 'from within VS'. Advantage: has to be done once only, works on commandline as well as in VS, works for everybody, is pretty flexible and extensible. For example you could override OuputPath based on whether or not a certain file is present on the system, or a certain environment variable, or hostname, etc.
The answer you gave also works, though is also not from within VS. But the biggest drawback for me is that if you give your project to someone else they won't have an overriden output path, i.e. such modifications cannot really be put in version control. Of course if that's what you are after than it's fine.
I compiled a .dll file in ironpython, by using the following code:
import clr
clr.CompileModules('C:/example.dll', 'C:/example.py')
It essentially compiles the .py file to .dll.
The only problem with this is that it creates a file with no information about the Company, Language, File Version etc. In fact the File Version is always: 0.0.0.0.
I was wondering if there is a way to at least alter the File Version (change it to something other than 0.0.0.0). I googled and found a similar topic in here on stackoverflow.
I tried three methods:
1) One with Visual Studio (File->Open-> find .dll, Edit->Add Resource->Version click New. In the new Version tab, change FILEVERSION and PRODUCTVERSION)
2) Another one by using the Change version 2012 application
3) And third one by using: Simple Version Resource Tool for Windows 1.0.10
None of them worked.
For some reason looks like the structure of the .dll assembly created with ironpython is different than the .NET one created with VB or C#.
Does anyone know how to change the File Version from 0.0.0.0 to something else?
Thank you.
You can use the pyc.py file packaged into IronPython to compile your file into a .dll file. The file is located in the directory IronPython 2.7\Tools\Scripts.
If we open pyc.py for editing, you'll see the different things it can do.
pyc: The Command-Line Python Compiler
Usage: ipy.exe pyc.py [options] file [file ...]
Options:
/out:output_file Output file name (default is main_file.<extenstion>)
/target:dll Compile only into dll. Default
/target:exe Generate console executable stub for startup in addition to dll.
/target:winexe Generate windows executable stub for startup in addition to dll.
#<file> Specifies a response file to be parsed for input files and command line options (one per line)
/file_version:<version> Set the file/assembly version
/? /h This message
EXE/WinEXE specific options:
/main:main_file.py Main file of the project (module to be executed first)
/platform:x86 Compile for x86 only
/platform:x64 Compile for x64 only
/embed Embeds the generated DLL as a resource into the executable which is loaded at runtime
/standalone Embeds the IronPython assemblies into the stub executable.
/mta Set MTAThreadAttribute on Main instead of STAThreadAttribute, only valid for /target:winexe
/file_info_product:<name> Set product name in executable meta information
/file_info_product_version:<version> Set product version in executable meta information
/file_info_company:<name> Set company name in executable meta information
/file_info_copyright:<info> Set copyright information in executable meta information
/file_info_trademark:<info> Set trademark information in executable meta information
Example:
ipy.exe pyc.py /main:Program.py Form.py /target:winexe
One thing that I personally like to do is move pyc.py from the Scripts folder to the IronPython folder along with my python file as well.
Assuming you also do this, you would open command prompt as administrator and navigate to the IronPython folder.
cd "Program Files (x86)\IronPython 2.7"
Then you would want to compile your python file as a .dll and set the file version using pyc.py. To do that, you're going to want to type in:
ipy.exe pyc.py /main:example.py /target:dll /file_version:0.0.0.1
If you want to add a company name, and other items as well, you simply have to pass those option to the pyc.py script.
ipy.exe pyc.py /main:Program.py Form.py /target:winexe /file_info_company:Company
Can I include an .exe file in another, and then run it from the outer program?
For instance, can I make a wget GUI by including it inside my program, or are my only options either using the including the source or supplying the wget binary together with my wrapper?
I am working on Windows and am looking for a solution in c/c++/c#
Sure you can.
The idea is to 'insert' the exe as a resource to you main application.
There is a link which explains how to compile resources into delphi exe. Its similar to VC++ or what ever...
I have written a small VB.NET simulation program that uses an XML file to configure the simulation. I want to include this file in the project build so that when the application is installed, there will be a default XML file in the required directory.
When I do the Project Publish (within VB 2010 Express), there is no option for including any extra data files in the process.
Is it possible to do this with VB 2010 Express ... or should I try some other project builder/installer.
Any pointers will be very much appreciated,
Regards,
Oliver
The option isn’t found in the publisher, it’s a property of the file itself: when you add a file to the project you can set its file properties in the property window (usually at the right-hand side of the screen, below the file browser).
There you can set its “Build Action” to “Content” and its “Copy to Output Directory” mode to “Copy if newer”.
I have a Visual Studio 2010 solution that is set to build in Debug x86. Visual Studio therefore sets the output path to \bin\x86\Debug, which seems logical enough.
The solution contains about 50 projects; the start-up project is a WCF project.
When I do a build, I would expect all output dlls to go to \bin\x86\Debug, as that is what is set in the project settings. But weirdly, I see dlls being created in bin\x86\Debug and in \bin. Why would Visual Studio put any dll in \bin if the output path is not set to that directory? It seems that all dlls go to \bin\x86\Debug, and all dlls except for the start-up project go into \bin. Any idea why it would do that? (We have other solutions that don't use WCF, and they don't have this problem.)
The other annoyance is if I run the service from Visual Studio and then try to access my service in a web browser, by going to http://localhost:1240/MyService.svc, it doesn't work, because the start-up project dll is missing from /bin. I therefore have to manually copy this one dll from \bin\x86\Debug to \bin, so that all dlls are found and the service runs normally. (We could of course add a custom post-build step that does the copy, but you'd think there'd be a better way!)
To those of you working on WCF projects, do you leave the output path at \bin\x86\Debug? (Perhaps there is a way to configure the service, eg in the web.config or .svc file, so that it knows the binaries are in \bin\x86\Debug instead of \bin?) Or do you change the output path to \bin so that you can run your service straight from Visual Studio?
If you open the property page of you WCF Hosting project and goto tab Build, under the Output section of this tag there is a textbox that contains the location of the output binaries.
For a WCF project the binaries should go to bin directory irrespective of the build type (suc as Debug, Release). Make sure this value has been configured correctly.
The value needs to be configured for each build type\configuration.
If you need, for whatever reason, different outputs from different build configurations in different folders you could specify them like you did first and use a post build event command line that copies from your specified output folder to bin.
Like:
- untested code -
COPY/Y "$(OutDir)\*.*" "$(SolutionDir)$(ProjectName)\bin\"
- untested code -