Automatic copy files to output during application building - c++

Question

There is Copy to Output Directory property for files in C# projects. But in VC++ projects it is absent. I know, that I can use Build events in VC++ and write there something like
xcopy /y /d %(FullPath) $(OutDir)
Is there a way to avoid the use of CMD (and other scripting methods)? Can msbuild do something to help in this case?

Answer 1

Can MSBuild do something to help in this case?
Using MSVC 2012, this worked for me:
Assumed you have a file "Data/ThisIsData.txt" in your c++ Project.
Unload the project (right click --> Unload Project).
Edit project XML (right click --> Edit .vcxproj)
Now you see the projects MSBuild file as XML in your editor.
Find "ThisIsData.txt". It should look something like:
<ItemGroup>
<None Include="Data\ThisIsData.txt" />
...
</ItemGroup>
Now add an other item group like this:
<ItemGroup>
<Content Include="Data\ThisIsData.txt">
<CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
</Content>
...
</ItemGroup>
Reload the project and build.
Your file "ThisIsData.txt" should get copied to $(OutDir)\Data\ThisIsData.txt.
Why duplicating the ItemGroup?
Well if you simply change the None include to a content include, the IDE does not seem to like it any more, and will not display it. So to keep a quick edit option for my data files, I decided to keep the duplicated entries.

Answer 2

In VS 2015 it is possible to give C projects the functionality that is in C#.
(Idea from building off of jochen's answer.)
Instead of adding another ItemGroup, modify the given itemgroup adding a CopyTo element. I.E, using his example, simply enhance the original entry to:
<ItemGroup>
<None Include="Data\ThisIsData.txt" />
<DeploymentContent>true</DeploymentContent>
<CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
...
</ItemGroup>
No other ItemGroup required. By adding the CopyTo element, you add an "Included In Project" property.

Answer 3

In Visual Studio 2017 you can do this in the IDE. I am not sure about earlier versions.
Simply add the file as an included project file so it shows in the Solution Explorer. Then right click on the file and select the Properties menu.
Change the Content to "Yes" and change the Item Type to "Copy file"
If you look at the changes it made to the project file you can see it added this:
<ItemGroup>
<CopyFileToFolders Include="Filename.txt">
<DeploymentContent>true</DeploymentContent>
<FileType>Document</FileType>
</CopyFileToFolders>
</ItemGroup>

Answer 4

It depends on what version of Visual Studio you are using. Format of VC++ project file in Visual Studio 2008 is not MSBuild and so using xcopy in PostBuildStep is a good choice.
VC++ project in Visual Studio 2010 has MSBuild format. Thus, there is functionality of MSBuild Copy task.
Below is a sample:
<Copy
SourceFiles="%(FullPath)"
DestinationFolder="$(OutDir)"
/>
If the destination directory does not exist, it is created automatically
An MSDN Copy task reference is here

Answer 5

Following henri-socha's answer about VS2015 (and probably VS2013 and VS2012, or anything using MSBuild style projects), the ItemGroup item type is important.
Specifically <Text> items do not seem to be copied, whereas <Content> items do.
So, for a project directory Data containing a text file ThisIsData.txt, this will create a subdirectory Data under the $(OutDir) directory and copy the file ThisIsData.txt from the project into it if it's newer:
<ItemGroup>
<Content Include="Data\ThisIsData.txt">
<CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
</Content>
</ItemGroup>
This won't, although it is what the Visual Studio IDE will insert if you add the text file to your project, and set the Content property to True.
<ItemGroup>
<Text Include="Data\ThisIsData.txt">
<DeploymentContent>true</DeploymentContent>
</Text>
</ItemGroup>
So in other words you need to add the file via the IDE to make it realise the file is included in the project (which adds <Text> tag ItemGroup), and then open the project in a text editor and add the <Content> tag ItemGroup to get it to do what you want.
I'm not sure what the <DeploymentContent> tag actually does. It may be a remnant since the only MSDN reference I could find considers it archived: https://msdn.microsoft.com/en-us/library/aa712517.aspx

Answer 6

In visual studio 2019 after setting the file as "Include in project" you can edit the properties an select as Item Type "Copy file" (as shown in https://i.stack.imgur.com/vac2b.png)
This avoids the manual vcxproj file edition.

Answer 7

You can specify copying in the project file as Jeff G answered in another question:
In the *.vcxproj file, change:
<Text Include="Filename.txt" />
to:
<Content Include="Filename.txt">
<CopyToOutputDirectory>PreserveNewest</CopyToOutputDirectory>
</Content>
Then in the *.vcxproj.filters file, change:
<Text Include="Filename.txt">
<Filter>Resource Files</Filter>
</Text>
to:
<Content Include="Filename.txt">
<Filter>Resource Files</Filter>
</Content>
where the <Text ...> tag is for specified text files (it'll be <Image ...> for image files etc.)

Answer 8

If it's a COM dll, you can add it to the root of your project, mark it as 'Content' and set copy to output directory to 'Always'. I had to do this for signature capture COM assembly.