Visual Studio loading the right (x86 or x64) dll!

Question

I'm working on visual studio in an x86. I would like to build my application for both x32 and x64. But i need to use the sqlite .net connector wich has a dll for x86 apps and another dll for x64 apps. How do i configure my visual studio to load a reference when my configuration is x64 and another when my configuration is x86?

Thanks, Richard.

Answer 1

in your project file in reference use an MSBUILD conditional

<Reference 
       Include="SomeAssembly86, Version=0.85.5.452, Culture=neutral, PublicKeyToken=41b332442f1101cc, processorArchitecture=MSIL"  
         Condition=" '$(Platform)' == 'AnyCPU' ">
      <SpecificVersion>False</SpecificVersion>
      <HintPath>..\..\Dependencies\SomeAssembly.dll</HintPath>
      <Private>False</Private>
    </Reference>
    <Reference 
         Include="SomeOtherAssembly, Version=0.85.5.999, Culture=neutral, PublicKeyToken=41b332442f1101cc, processorArchitecture=MSIL" 
         Condition=" '$(Platform)' == 'x64' ">
      <SpecificVersion>False</SpecificVersion>
      <HintPath>..\..\Dependencies\SomeOtherAssembly.dll</HintPath>
      <Private>False</Private>
    </Reference>

Answer 2

This slightly simpler answer than Preet Sangha's will not generate a warning when the project is loaded and only the conditionally accepted dll will appear in the Solution Explorer. So, overall, the appearance is cleaner, although more subtle. (This was tested in Visual Studio 2010.)

<Reference Include="p4dn" Condition="$(Platform) == 'x86'">
  <HintPath>..\..\ThirdParty\P4.Net\clr4\x86\p4dn.dll</HintPath>
</Reference>
<Reference Include="p4dn" Condition="$(Platform) == 'x64'">
  <HintPath>..\..\ThirdParty\P4.Net\clr4\x64\p4dn.dll</HintPath>
</Reference>

Answer 3

You can also build your application for "Any CPU" and dynamically choose which DLL to load.