I got a CSharp project which uses a generated source file. The sourcefile is generated as a PRE_BUILD command.
add_custom_command(
TARGET ServiceTest PRE_BUILD
COMMAND cmd.exe /c "#echo content of file > \"$(ProjectDir)sourcefile.cs\""
)
The file generated is included this way and located in the binary tree of that project:
set(SOURCE_FILE "${CMAKE_CURRENT_BINARY_DIR}/sourcefile.cs")
set_source_files_properties(${SOURCE_FILE} PROPERTIES GENERATED true)
add_executable(ServiceTest <Some other Sources> ${SOURCE_FILE})
My problem is that I can't seem to visually get the generated file into the explorer in Visual Studio. I tried reloading the project and checking the projectfile itself. Everything looks ok. The only difference is that the root folder of the source is different. I also thought of cmake not finding the file so I also tried an empty initial file. It does not work.
Does anyone have an idea how I could get Visual Studio to immediately show the generated Source?
I'm new to Cmake and I have a main output named TARGET. I'm trying to add custom target named _COPY_ASSETS_TARGET as a dependency to main TARGET. I want TARGET to rebuild _COPY_ASSETS_TARGET automatically if there's any change in _COPY_ASSETS_TARGET. _COPY_ASSETS_TARGET should depend on change in folders.
Here's the code I tried to implement:
if (NOT TARGET ${_COPY_ASSETS_TARGET})
add_custom_target(${_COPY_ASSETS_TARGET})
add_dependencies(${_ARGS_PROJECT_TARGET} ${_COPY_ASSETS_TARGET})
set_property(TARGET ${_COPY_ASSETS_TARGET} PROPERTY FOLDER "Targets")
endif()
add_custom_command(TARGET ${_COPY_ASSETS_TARGET}
${_COMMANDS}
VERBATIM
)
I'm trying visual studio to debug and if I rebuild _COPY_ASSETS_TARGET then only I can see the updated output. I would like to know how I can link my folders to _COPY_ASSETS_TARGET so that TARGET automatically builds new code
add_custom_command(OUTPUT ${OUTPUT_DIRECTORY} ${_COMMANDS}
DEPENDS ${DEPENDENT_DIRECTORY}
VERBATIM
)
add_custom_target(${_COPY_ASSETS_TARGET} ALL
DEPENDS ${OUTPUT_DIRECTORY}
)
add_dependencies(${_ARGS_PROJECT_TARGET} ${_COPY_ASSETS_TARGET})
set_property(TARGET ${_COPY_ASSETS_TARGET} PROPERTY FOLDER "Targets")
I found this to be working for my project
I have a standard (?) project (vcxproj) with multiple .asm files in Visual Studio 2019 Community 16.6.
In Configuration properties->MASM I have not set anything
Still the .lst file is created for the project. So, there must be some "hidden" defaults somewhere. However, even if I set them explicitly - there is no setting as to for which .asm file the listing file should be created.
I could not find anything as to the default implicit behavior / explicit behavior of choice of one .asm file over another. Logically, it would be the file containing the entry point in x86-32bit "end main", but that's not the case. Also, is there a way to generate .lst next to the .obj file for all .asm files in a given project.
right for me:
Open Project ,
Create a file (*).asm AND write code
Go to Property of Project: ( note platform for correct in first line - for me: WIN32 )
Go to Microsoft Macro Assembler - Listing file - set Yes(/Sg) Enable Assembly Generated Code Listing
and set $(ProjectName).lst in Assembled Code Listing File
Go to property of file (*).asm do the same thing as above...
Thanks for watch
When I build a Visual Studio project, the executable is written to the output directory specified in the projects Property Page.
I have a project that has some extra files (e.g., .ini file) that are used by the program.
How can I configure the project to copy the file to the output directory so that when the program runs, it has a copy of the other file in its CWD?
I checked the Property Page of the file and there was nothing useful other than an option to exclude it from the build (which is disabled), and the custom-build-tool command is empty (plus it is a plain-text file that does not need any processing).
For copying a files to the output directory in Visual Studio 2003 you could use Post-Build event:
Right click on the project->Properties
Common Properties->Build Events
Set Post-Build Event Command Line to:
xcopy /y $(ProjectDir)my_file.ini $(ProjectDir)$(OutDir)
OK and build!
Please try select the file in Solution Explorer. Then you should be able to see its properties in Properties window (press F4 if it is not visible). You will find there two properties:
"Build Action" and
"Copy to Output Directory"
Set "Build Action" to "Content", and then - select an appropriate value for "Copy to Output Directory" setting.
File properties window with "Build Action" and "Copy to Output Directory" settings
If the way above doesn't work for you, please read this post "Copy to output directory issue with .inf file". And have a look at this one then "Visual Studio: default build action for non-default file-types"
While I was searching the file’s Property Page for a build-action field, I had a thought: set the custom build step to copy the file (manually). This turned out to be easier than I thought. I had figured it would require using cmd or other external executable (xcopy, robocopy, etc.), but that is not necessary.
I set the Custom Build Step as follows:
Command Line : copy $(InputFileName) $(OutDir)
Description : Copying foobar...
Outputs : $(InputFileName)
Setting the outputs field (correctly) was critical in order to prevent VS from always thinking the project is out of date and requiring to be rebuilt (I’m not certain if it needs to be prefixed with $(OutDir)\).
It is reflected in the Output window as such:
Copying foobar...
1 file(s) copied.
Compiling resources...
Linking...
For VS 2017 the command Dmitry Pavlov posted would be the following:
xcopy /y "$(ProjectDir)my_file.ini" "$(OutDir)"
Quotes are important in case there are spaces in the path to the project directory.
Expanding on Synetech's answer.
In VS2019 right click the file you want to copy in the Solution Explorer and select Properties. Then under General >> Item Type change to Copy File then hit Apply.
You now should have UI fields in the Properties Page for Destination etc.
In case this helps anyone, I needed to copy the output dll of the project i was building into another project.
xcopy /y "$(ProjectDir)$(OutDir)$(TargetName)$(TargetExt)"
"C:\Application\MyApplicationName\bin\x86\Debug"
/y = overwrite file if already exists
$(ProjectDir) = location on your machine where the project lives
$(OutDir) = is where your current build setup outputs the build
$(TargetName) = What the project being built is set to be called. Ex: XXX of XXX.dll
$(TargetExt) = the extension of the build Ex: .dll of XXX.dll
"C:/..../x86/Debug" is the location to copy to.
You need the extra $(OutDir). Otherwise, in the rebuild/clean step it will throw away your source.
CommandLine : copy "$(SolutionDir)last-script.js" "$(TargetDir)Debug"
Outputs : $(TargetDir)Debug\last-script.js
Improving Synetech
answer :
In VS 2013 C++ project Command Line : copy %(Identity) $(OutDir) Description : Copying foobar... Outputs : %(Identity)
It works , But it leads to circular dependency , i.e. it will be executed each time you demand increamental build, no meter it has been already copied.
To solve this , you can add that item at target folder, change path to $(OutDir), and use that in first added item as Output. Drawback - two items with similar name are in solution.
Also usefull xcopy with /d /y parameters in postbuild - copy only if target file date is older.
You could also after the unload the project (Right click on the project >> Unload Project) add the following inside an existent <ItemGroup> tag:
<Content Include="..\..\Config\db.config">
<Link>Config\db.config</Link>
<CopyToOutputDirectory>Always</CopyToOutputDirectory>
<SubType>Designer</SubType>
</Content>
In this case it will grab the db.config file from 2 folders up and put it in the output folder(by default bin/Debug) after creating a Config folder with the db.config file inside
I'm about to start doing some benchmarking/testing of our builds, and I'd like to drive the whole thing from a command line. I am aware of DevEnv but am not convinced it can do what I want.
If I could have a single file built within a single project, I'd be happy.
Can this be done?
The magical incantation is as follows. Note that this has only been tested with VS 2010 - I have heard this is the first version of Visual Studio with this capability:
The Incantation
<msbuild> <project> <settings> <file>
Where
msbuild is a path to MSBuild.exe. Usually this should be set up for you by the VS2010 bat file so the right one will end up in your PATH, but if not I found one at C:\Windows\Microsoft.NET\Framework64\v4.0.30319\msbuild.exe
project is the path to the vcxproj file within which your source file resides.
settings include the following:
/p:Configuration="Debug" // or whatever your Configuration is
/p:Platform=x64 // or x86
/t:ClCompile // to specify specifically you're looking to compile the file
file is actually another setting:
/p:SelectedFiles="path_to_file"
Notes
For <project> I had to specify a project (vcxproj) file instead of a solution (sln) file. The sln I would have used has multiple projects within it, so there would have been extra work to go that route anyhow (if it can even be done).
For the /p:Platform=x64 setting, there are several environment variables that pivot on what platform you are targeting (x64 v. x86) so make sure you set those up properly via Visual Studio's vcvarsall.bat.
Regarding path_to_file in the SelectedFiles parameter, this path must be the path as specified in the project file. If the path does not match the path used in the project file to reference the source, it doesn't seem to work.