Re: Warning

From:
Michael Doubez <michael.doubez@free.fr>
Newsgroups:
comp.lang.c++
Date:
Mon, 8 Mar 2010 07:23:10 -0800 (PST)
Message-ID:
<0c4c83bc-bed1-459d-b884-4c39b240485b@q15g2000yqj.googlegroups.com>
On 8 mar, 15:49, "Leigh Johnston" <le...@i42.co.uk> wrote:

"Michael Doubez" <michael.dou...@free.fr> wrote in message

news:64baf247-ed93-400a-b1f5-be2c5119abb1@d2g2000yqa.googlegroups.com...

On 8 mar, 13:48, "Leigh Johnston" <le...@i42.co.uk> wrote:

"Michael Doubez" <michael.dou...@free.fr> wrote in message

news:bffc3535-8ad3-4816-b3cf-83f72c78e556@j27g2000yqn.googlegroups.com.=

...

On 8 mar, 12:01, "Leigh Johnston" <le...@i42.co.uk> wrote:

"Michael Doubez" <michael.dou...@free.fr> wrote in message

news:d962d8ed-d69e-4b5f-a224-bebd8dcafb83@e7g2000yqf.googlegroups.co=

m...

On 5 mar, 11:25, "Leigh Johnston" <le...@i42.co.uk> wrote:

"Michael Doubez" <michael.dou...@free.fr> wrote in message
news:cdb20f88-8668-4092-972e-85a119dff724@j27g2000yqn.googlegroup=

s.com...

On 4 mar, 19:37, James Kanze <james.ka...@gmail.com> wrote:

[snip]

Deriving from classes which weren't designed to be bases (suc=

h

as the standard containers) is generally a bad idea.


That's because, IMO the standard containers usually have a
complete
interface and there is no need to.


The examples I gave where for the case where an interface needs =

to

be
augmented but you use the word "usually" which is fine: "usually=

"

is
not
"always".


But the corollary is that is seldom useful or a good design decis=

ion

to inherit from them. A complete interface means that functions o=

r

composition should be used; AFAIS there are three cases:
 - the extended part is stateless: a function should be used
 - the extended part keeps associated state:
    * the extended class should be somewhat notified of modif=

ication

but the base class is not designed that way and composition shoul=

d

be
used (I exclude an external notify modification system)
    * the extended class doesn't need to be notified, both
information should be composed as a pair in another structure
    * the extended structure exploits the internal of the con=

tainer:

this is not portable, even across the same version of the compile=

r.

Garbage.


Really ?
A minimal complete interface is IMO an essential value for general
purpose class design (I am not talking about classes implementing
specific use cases).

The "is-a" relationship is well defined


I am curious to hear your definition.

and perfectly fine.


I find it ambiguous: it depends on what you mean by "is".

In fact, I find it useful only in defining what is-not (i.e. what
should be composed and not inherited).


Example of "is-a":

struct widget
{
  colour background;
  virtual void draw(device_context& dc) const
  {
    dc.fill_rect(client_rect(), background);
  }
  colour get_background() const { return background; }

};

struct label : widget
{
  virtual void draw(device_context& dc) const
  {
    widget::draw(dc); // erases background in background colour
    dc.draw_text(0, 0, get_background() ^ -1, label_text()); // dr=

aw text

in
inverted colour
  }
};

label "is-a" widget, i.e. it inherits a widget's ability to fill in it=

s

background and this behaviour is optional (label does not have to call
widget::draw()). label also inherits widget's background colour and=

 can

query it for use in its own drawing code.


If I have a Spacer widget that doesn't draw anything, the background
member is useless.

And the name widget (WIndow gaDGET) is not really a thing, it is more
a base class for elements contained within a window. To me, it looks
like something you are forced into by strong typing rather than a
design decision.

"is-a" is related to LSP:

struct window
{
  std::vector<widget*> widgets;
  device_context dc;
  void draw()
  {
    for (auto i = widgets.begin(); i != widgets.end(); ++i)
      i->draw(dc);
  }
}

i.e. if LSP is adhered to then a label can be passed to any function
which
accepts a widget reference/pointer. This is the essence of the "is-=

a"

relationship.


Well, is-a yields a correct program only if it preserves LSP. Although
there are some cases where LSP is not preserved: if I make Integer and
String subtypes of AdditiveType, Integer addition is symmetric but
String addition/concatenation is not although math tells us that '+'
is reserved for symmetric operations.

From the examples you gave, I see that, for you, is-a is thought in
terms of interface and polymorphism, not in terms of subtyping.

Now, if I have a class Mammal:
class Mammal
{
 public:
   Mammal(unsigned nb_breast):nb_breast(nb_breast){}

   unsigned nbBreast()const{ return nb_breast; };
 protected:
    unsigned nb_breast;
};

If I define cat is-a mammal:
class Cat: public Mammal
{
 public:
 enum Type{ /* type of cat */};

 Cat(Type type): Mammal(catType2NbBreast(type)){}

 void mastectomy(unsigned nb_breast_removed)
 {
  assert( nb_breast_removed <= nb_breast);
  nb_breast -= nb_breast_removed;
 }
};

Here, I have a is-a relationship without talking about LSP or
polymorphism.

As I said elsewhere, the fact that C++ implements (dynamic)
polymorphism in terms of inheritance doesn't help. Well, it couldn't
do it another way, now, could it ?


Your example is just an example of ordinary derivation, I use derivation =

for

non-polymorphic classes also. Both the "is-a" relationship and LSP do =

not

require polymorphism to be valid I never said they did, I just happened t=

o

give an example which was polymorphic.


LSP does concern polymorphism otherwise it is useless. I merely gave
an example of is-a relationship without polymorphism to show LSP is
not part of "the essence of 'is-a' relationship".

"is-a" relationship holds true if LSP holds true. Your example does no=

t

violate LSP, I can pass a Cat to a function requiring an Mammal reference
and calling the nbBreast() member function will work.


Yes but LSP is not involved in any way in my example.

I think we have lost the trail of the discussion somewhere. My claim
was that is-a relationship is ambiguous and is usually assimilated to
polymorphism - which it is not.

And you are proving me right: the first example you give is pure
polymorphic behavior and you define it in relation to LSP.

Unless you are claiming that a is-a relationship is necessarily
polymorphic; in which case I will ask: isn't my example a is-a
relationship ?

--
Michael

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