How to force an application to use .NET 3.5 or above?

Question

Our application is built with VS 2008, uses Linq and has Target Framework set to .NET Framework3.5.

It works OK when only .NET 3.5 or 4 is installed on the machine.

However, on machines where both .NET 2 (or 3.0) and .NET 4 are installed, the application is loaded with .NET 2, and crashes when Linq is accessed, as it looks for the .NET 3.5 libraries.

Using the tag in app.config doesn't seem to help, as it specifies the CLR version, which is 2 in case of .NET 3.5.

Note that our installation verifies that .NET 3.5 or upper is installed.

Is there a way to tell the application to load:

• the highest CLR it finds, or
• CLR 4 if it is installed, and CLR 2 if CLR 4 is not installed, or
• CLR 2 if .NET 3.5 is installed and CLR 4 if .NET 3.5 is not installed

(Note that similar question is left unanswered in the Community Content section of the Element documentation )

Answer 1

Forming the question led me to the answer. As mentioned in the Element documentation ,

When multiple versions of the runtime are supported, the first element should specify the most preferred version of the runtime, and the last element should specify the least preferred version.

So the way to achieve the second option ("CLR 4 if it is installed, and CLR 2 is CLR 4 is not installed") is to reverse the order of the elements in app.config:

<?xml version="1.0" encoding="utf-8" ?>
<configuration>
    <startup useLegacyV2RuntimeActivationPolicy="true">
        <supportedRuntime version="v4.0"/>
        <supportedRuntime version="v2.0.50727"/>
    </startup>
</configuration>

This way, .NET 4 will be loaded if it is installed, and an earlier version will be loaded if not.

Answer 2

Another useful link is this page on MSDN. This shows all of the values required in app.config if you want to just target the client profile or if you require the full profile.

Answer 3

.NET Framework 3.0和3.5版使用CLR的2.0.50727版。

Answer 4

如果您有一个非托管EXE调用.NET DLL，您还需要创建一个foo.exe.config文件，其中包含上面的<startup>...块。