Can the DebuggerDisplay attribute be applied to types one doesn't own?

Example of setting DebuggerDisplay for a foreign type (System.Collections.Generic.KeyValuePair<TKey,TValue>) add the following to AssemblyInfo.cs:

using System.Collections.Generic;
using System.Diagnostics;

[assembly: DebuggerDisplay("[Key={Key}, Value={Value}]", Target = typeof(KeyValuePair<,>))]

(Tested in VS2015)

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Edit 2020:

Was not able to reproduce the above for KeyValuePair<,> in VS2019 but it seems to be related to KeyValuePair<,>.

For private member of non owned types try something like this

ClassLibrary1:

//using System.Diagnostics;

namespace ClassLibrary1
{
    //[DebuggerDisplay("Foo.Bar={Bar}")] // works too for types you own
    public class Foo
    {
        private int Bar = 42;
    }
}

ConsoleApp1:

using System.Diagnostics;
using System.Reflection;
using ClassLibrary1;

[assembly: DebuggerDisplay("Foo.Bar={FooDebuggerDisplay.Bar(this)}", Target=typeof(Foo))]

class FooDebuggerDisplay
{
    public static int Bar(Foo foo) => (int)foo.GetType().GetField("Bar",BindingFlags.Instance|BindingFlags.NonPublic).GetValue(foo);
}

namespace ConsoleApp1
{
    class Program
    {
        static void Main(string[] args)
        {
            var foo = new Foo();
            Debugger.Break();
        }
    }
}

(Tested in VS2019)

Yes. In fact, Microsoft was so nice as to make this a built-in option in Visual Studio.

Look up the "My Documents\Visual Studio 20XX\autoexp.cs" for some examples of how you apply the DebuggerDisplay attribute to types that are foreign to your assembly. Then, add some of your own, recompile it and replace the autoexp.dll, and restart Visual Studio. It should Just Work.

For reference, see the yellow "Note" paragraph in this MSDN article

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Alternatively: I'm the creator of a purchasable extension to Visual Studio that enables doing this with a lot less fuss, without even needing to have to stop the debugging session.