Why can't I use a try block around my super() call?

Unfortunately, compilers can't work on theoretical principles, and even though you may know that it is safe in your case, if they allowed it, it would have to be safe for all cases.

In other words, the compiler isn't stopping just you, it's stopping everyone, including all those that don't know that it is unsafe and needs special handling. There are probably other reasons for this as well, as all languages usually have ways to do unsafe things if one knows how to deal with them.

In C# .NET there are similar provisions, and the only way to declare a constructor that calls a base constructor is this:

public ClassName(...) : base(...)

in doing so, the base constructor will be called before the body of the constructor, and you cannot change this order.

It's done to prevent someone from creating a new SecurityManager object from untrusted code.

public class Evil : SecurityManager {
  Evil()
  {
      try {
         super();
      } catch { Throwable t }
      {
      }
   }
}

I know this is an old question, but I liked it, and as such, I decided to give it an answer of my own. Perhaps my understanding of why this cannot be done will contribute to the discussion and to future readers of your interesting question.

Let me start with an example of failing object construction.

Let's define a class A, such that:

class A {
   private String a = "A";

   public A() throws Exception {
        throw new Exception();
   }
}

Now, let's assume we would like to create an object of type A in a try...catch block.

A a = null;
try{
  a = new A();
}catch(Exception e) {
  //...
}
System.out.println(a);

Evidently, the output of this code will be: null.

Why Java does not return a partially constructed version of A? After all, by the point the constructor fails, the object's name field has already been initialized, right?

Well, Java can't return a partially constructed version of A because the object was not successfully built. The object is in a inconsistent state, and it is therefore discarded by Java. Your variable A is not even initialized, it is kept as null.

Now, as you know, to fully build a new object, all its super classes must be initialized first. If one of the super classes failed to execute, what would be the final state of the object? It is impossible to determine that.

Look at this more elaborate example

class A {
   private final int a;
   public A() throws Exception { 
      a = 10;
   }
}

class B extends A {
   private final int b;
   public B() throws Exception {
       methodThatThrowsException(); 
       b = 20;
   }
}

class C extends B {
   public C() throws Exception { super(); }
}

When the constructor of C is invoked, if an exception occurs while initializing B, what would be the value of the final int variable b?

As such, the object C cannot be created, it is bogus, it is trash, it is not fully initialized.

For me, this explains why your code is illegal.