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Bridging Static and Dynamic Typing

1 reply on 1 page. Most recent reply: Apr 17, 2006 5:14 AM by Achilleas Margaritis

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Christopher Diggins

Posts: 1215
Nickname: cdiggins
Registered: Feb, 2004

Bridging Static and Dynamic Typing (View in Weblogs)
Posted: Apr 13, 2006 6:59 AM
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Michael Feathers recently suggested that maybe static typing is a form of bad coupling which we could do without. Well in C++ at least you can bridge static typing and dynamic typing.
Okay I lured you in, but I'll have to admit what I am about to propose here isn't elegant. C++ support for both generic programming and dynamically typed programming is extremely rudimentary, however I think there is something to take away.

The bridge I am talking about is as follows:

  template<typename T = ootl::object>
  class priority_queue {
     void push(T value, int priority);
     T pop();
     bool is_empty();
In this example, ootl::object is my dynamically typed variant class which is available at but you can just as easily use boost::any (which is slower, but perhaps a bit safer).

So what do you think? Dynamically typed, or statically typed, its your choice when you use templates.

Achilleas Margaritis

Posts: 674
Nickname: achilleas
Registered: Feb, 2005

Re: Bridging Static and Dynamic Typing Posted: Apr 17, 2006 5:14 AM
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Coding a Variant class in C++ is no problem at all. Unfortunately, C++ does not offer 'variant' semantics, so it takes more code to use a 'variant' instance properly. If a variant is used in expressions, then operators can be passed in the underlying object as messages (i.e. virtual methods). But there is no way to call a method of an object of a variant instance without casting the variant to a type, which makes the whole issue of using 'variant' more of a problem than a solution.

In other words, one can happily do in C++ the following:

var x = 5;
x += 1;
cout << x;
var y = x;
y = "the quick brown fox";

But the following is not possible:

var x = new foo;

unless one does this:

foo *f = (foo *)x;

which makes using 'var' reduntant, since we are going to type 'foo' anyway. Not only that, but it makes it very dangerous, because if I change 'foo' to 'bar' and then I forgot to update references to 'foo' in other places, then my program will certainly crash.

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