Question

I do serialization via a serialization constructor like this:

private MyClass(SerializationInfo info, StreamingContext c)
{
   try
   {
      MyIntVar = info.GetInt32("MyIntVar");
   }
   catch(Exception)
   {
      Trace.WriteLine("Exception occured! Setting default value.");
      MyIntVar = 4711;
   }
}

What I’m now trying to achieve is to trace the name and path of the file that is beeing serialized when this exception occures.

Something like:

if( c is file)
{
   Trace.WriteLine("Don't bother, I proceed anyway, but maybe you should repair the file " + FilePath);
}

So I have two questions concerning this:

1. How can I determine that the current serialization context is a file?
2. How can I get the corresponding file name and path of this file?

Answer 1

The only way you can do that is if you’ve created the StreamingContext yourself, and have made some additional information available via the .Context property. For example:

var ctx = new StreamingContext(StreamingContextStates.File, "SomeFileName");
//                                                          ^^^^ = context
var serializer = new BinaryFormatter(null, ctx);
// then use serializer.Serialize / .Deserialize

and then in your constructor or callback, access it:

bool isFile = (c.State & StreamingContextStates.File) != 0;

string filename = c.Context as string;
if(filename != null) {
   // ...
}

Actually, string is pretty ambiguous – I would advise using your own custom context-type that can’t ever be confused for something else. For example:

var ctx = new StreamingContext(StreamingContextStates.File,
    new MyStreamingContext { File = "SomeFile" });
...
class MyStreamingContext {
    public string File {get;set;}
}
...
var context = c.Context as MyStreamingContext;
if(context != null) {
    string file = context.File;
    // ...
}