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Home » Extra libraries, Code snippets, applications etc. » C++ language problems and code snippets » Basic library interface and separation problem
Basic library interface and separation problem [message #21860] Mon, 08 June 2009 15:51 Go to next message
janwilmans is currently offline  janwilmans
Messages: 51
Registered: January 2007
Member

Most of my projects aren't very well designed, when I even bother to separate different functionality into different libraries, I often use internal knowledge of the underlying library, even without thinking about it.

For my most recent project, I wanted to have proper separation and I ran into the following issue:

LibA
|
LibB
|
App

suppose we have library 'LibA' and 'LibB', LibB uses LibA and exposes part of LibA's functionality to an application using LibB.

LibA contains class A,

class A {
enum a_attrs { foo, bar, foobar }
static void setAttr(a_attris an_attr);
}

LibB contains class B,

class B {
static void setAttr(a_attris an_attr)
{
A::setAttr(an_attr);
}
}

This is a very bad solution for me, because I want to be able to replace LibA with LibA2 later, _and_ I don't want the App to even be aware of the existence of LibA.

Also, as it is now the App will include the headers of class A, indirectly, because App will include the header of class B.

So now, App has a source code level dependency on LibA .

One way to solve this would be to give class B it's own enum

class B {
enum b_attrs { b_foo, b_bar, b_foobar }
static void setAttr(b_attrs an_attr);
}

void B::setAttr(b_attrs an_attr)
{
switch(an_attr)
{
case b_foo:
A::setAttr(foo);
break;
case b_bar:
A::setAttr(bar);
break;
case b_foobar:
A::setAttr(foobar);
break;
default:
break;
}
}

But when I add an attribute to class A, in would also have to edit class B to add the attribute there as well.

Any thought on how to solve this neatly ?

Greetings,

Jan








Jan (skyhawk)

[Updated on: Mon, 08 June 2009 15:52]

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Re: Basic library interface and separation problem [message #21873 is a reply to message #21860] Mon, 08 June 2009 22:57 Go to previous messageGo to next message
koldo is currently offline  koldo
Messages: 2988
Registered: August 2008
Veteran
Hello skyhawk

Sorry, with so many A, and B, and foo, and bar, ..., I am lost!.

For me, when I see that I have two classes that share things I have three options:
- Redesign them dividing the functionalities so that finally I have different classes
- Make a common class with common funcionalities and derive both from this common class.
- Join them in one class.

You have to know how to cut and remix the classes. Or put a more defined sample and we can tell you our opinion.

Best regards
Koldo


Best regards
IƱaki
Re: Basic library interface and separation problem [message #21876 is a reply to message #21873] Tue, 09 June 2009 00:25 Go to previous messageGo to next message
Didier is currently offline  Didier
Messages: 486
Registered: November 2008
Location: France
Senior Member
Usualy what is used for this is: "dependancy inversion"
In other words: make an interface (it will be the API) and make both librairies depend on it (solution 2 of koldo).

The result is B does not depend on lib A although it uses it !
You can now freely change lib A or B, as long as you keep the API stable. Smile

This is what APIs are made for Cool

Re: Basic library interface and separation problem [message #21877 is a reply to message #21876] Tue, 09 June 2009 08:22 Go to previous message
janwilmans is currently offline  janwilmans
Messages: 51
Registered: January 2007
Member

Koldo, sorry for the confusion, what I was looking was that:

When two classes need to share more then just basetypes, in my opinion you have two options:

- make one class depend on the other
- create a third class that serves as 'common' API for both.

I my case, because the two class are in separate libraries and I do not want source code dependencies between them; I've done what Didier and you suggested, created a third library that is shared/depended on by the other two which contains a class that serves as common API for both libraries.

Anyway, thanks for your swift responses!

Jan


Jan (skyhawk)
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