Use 32bit “Program Files” directory in msbuild

Question

In 64 bit versions of windows, 32 bit software is installed in "c:\program files (x86)". This means you cannot use $(programfiles) to get the path to (32 bit) software. So I need a $(ProgramFiles32) to overcome this in my MSBuild project. I don't want to change the project depending on the os it is running on.

I have a solution that I will post, but maybe there is a easier/better way.

Answer 1

I think a slighly more reliable way is to grab the Environment variable "ProgramFiles(x86)". In a 64 bit process on Windows this will point to the 32 bit program files directory. It will be empty on a 32 bit version of windows and I believe on a wow64 process

I ran into virtually same problem recently with some PowerShell scripts. I wrote a blog entry on how a worked around the program files directory issue. Different language obviously but it may help you out.

http://blogs.msdn.com/jaredpar/archive/2008/10/21/program-files-i-just-want-the-32-bit-version.aspx

Answer 2

In MSBuild 4.0, $(MSBuildProgramFiles32) will give you the 32-bit Program Files directory.

Answer 3

I stumbled across this question trying to find a generic way in MSbuild to see if it was a 32- or 64-bit os. In case someone else also find this, I used the following:

<PropertyGroup>
  <OSBits Condition="$(ProgramW6432) != ''">x64</OSBits>
  <OSBits Condition="$(OSBits) == ''">x32</OSBits>
</PropertyGroup>

Apparently %ProgramW6432% is only set on 64-bit systems.

Answer 4

Try "$(MSBuildExtensionsPath32)\.."

Answer 5

If you run the 32-bit version of the Visual Studio tools (especially in VS2012, there are like 3 different command prompts you can choose from), $(ProgramFiles) points to "Program Files (x86)"

Answer 6

My solution is to look whether "c:\program files (x86)" exists, if it exists, asume this is a 64 bit os. Otherwise use the normal program files directory:

<PropertyGroup>
  <ProgramFiles32 Condition="Exists('$(PROGRAMFILES) (x86)')">$(PROGRAMFILES) (x86)</ProgramFiles32>
  <ProgramFiles32 Condition="$(ProgramFiles32) == ''">$(PROGRAMFILES)</ProgramFiles32>
</PropertyGroup>

I can use it like this

<Exec WorkingDirectory="src\app1" Command='"$(ProgramFiles32)\doxygen\bin\doxygen" Doxyfile' />