Find out the "Bit"ness of the current OS in MSBuild

Question

I have a build script that needs to hard code a path to an executable. The path is:

• C:\Program Files\Microsoft Visual Studio 9.0\SmartDevices\SDK\SDKTools\cabwiz.exe

This has worked fine, but now I am running on a 64 bit OS (but my coworker and build server are on 32 bit still).

I need the path to be this for me:

• C:\Program Files (x86)\Microsoft Visual Studio 9.0\SmartDevices\SDK\SDKTools\cabwiz.exe

But use the normal path for the others.

Here is how I set it up:

<PropertyGroup>
    <CabWiz>"C:\Program Files\Microsoft Visual Studio 9.0\SmartDevices\SDK\SDKTools\cabwiz.exe"</CabWiz>
</PropertyGroup>

Is there a condition I can put on that so that I can set it if the OS (not the current build configuration) is 64 bit?

Answer 1

On a 64-bit OS, the following variables are defined:

ProgramFiles=C:\Program Files
ProgramFiles(x86)=C:\Program Files (x86)

So just test for ProgramFiles(x86) and if it's empty, use ProgramFiles.

Answer 2

If you're always running the 32-bit version of MSBuild, regardless of the platform, then it's easy: just substitute '$(ProgramFiles)' for 'C:\Program Files'. Whether on a 32-bit or 64-bit OS, '$(ProgramFiles)' should expand to the correct folder location (the location of all 32-bit programs).

If you're running the 64-bit version of MSBuild on 64-bit platforms (which is unlikely), then it gets a bit trickier. The '%ProgramFiles(x86)%' environment variable would seem to be what you want, but good luck dealing with those parentheses. Easier would probably be to use the '%PROCESSOR_ARCHITECTURE%' environment variable in a condition.