C# – Will code in a Finally statement fire if I return a value in a Try block

The big difference is that try...catch will swallow the exception, hiding the fact that an error occurred. try..finally will run your cleanup code and then the exception will keep going, to be handled by something that knows what to do with it.

The finally statement is executed, but the return value isn't affected. The execution order is:

1. Code before return statement is executed
2. Expression in return statement is evaluated
3. finally block is executed
4. Result evaluated in step 2 is returned

Here's a short program to demonstrate:

using System;

class Test
{
    static string x;

    static void Main()
    {
        Console.WriteLine(Method());
        Console.WriteLine(x);
    }

    static string Method()
    {
        try
        {
            x = "try";
            return x;
        }
        finally
        {
            x = "finally";
        }
    }
}

This prints "try" (because that's what's returned) and then "finally" because that's the new value of x.

Of course, if we're returning a reference to a mutable object (e.g. a StringBuilder) then any changes made to the object in the finally block will be visible on return - this hasn't affected the return value itself (which is just a reference).