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Modified & Improved wxWidgets Project Wizard

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Pecan:

--- Quote from: Biplab on February 13, 2007, 05:28:37 pm ---By default Debug Target uses Debug wx lib with __WXDEBUG__ for all. But if it has to disable it then which library should be set default to for GCC?  :)

--- End quote ---

For gcc, NOT using __WXDEBUG__ in a debug target is simply a release target with the -g flag set. ie, just don't append the "d" to any filenames and don't set the -D__WXDEBUG__.

It would also be nice if "Enable all compiler warnings" was set. But I won't quibble over this.

What we're after here is a debuggable CodeBlocks (or any wxwidgets target) with debug symbols linked in, but a wxWidgets with NO debug symbols linked in.

This way, when you're stepping trough CodeBlocks (or any wxwidgets target), you won't accidently step into the wxWidgets can of worms. Accidently stepping into wxWidgets can take a major effort to get out, especially since GDB "stepout" doesn't seem to work.

It also means that the user can debug CodeBlocks (or any wxwidget target) without having to build/load/maintain a debug version of wxWidgets.

Having to tell a new plugin builder that s/he first has to build and deal with a debug version of wxWidgets in order to debug a simple plugin is not a pleasant thought.

TDragon:

--- Quote from: Biplab on February 13, 2007, 05:28:37 pm ---By default Debug Target uses Debug wx lib with __WXDEBUG__ for all. But if it has to disable it then which library should be set default to for GCC?  :)

--- End quote ---
The standard, non-debug library (i.e. libwxmsw26u.a for monolithic-unicode).

So if my suggested checkbox is checked, nothing changes: you get a standard debug target linking to the debug wx libs with __WXDEBUG__ defined. If the box isn't checked, the debug target uses the release wx libs and doesn't define __WXDEBUG__ (but still uses -g). Then you just make the box checked by default when using MSVC, and unchecked by default when using GCC.

Biplab:
First of all my apologies to Pecan and TDragon for this late reply. I got busy with my studies and couldn't touch this.

I have made the following changes. (Screenshot has been removed)

If "Don't use __WXDEBUG__ and Debug wxWidgets lib" is selected, wizard will-

* Not use debug lib of wx
* Will not use __WXDEBUG__
So the generated debug target will have -g option with release wx lib, but release target will remain as usual.

I've selected this option to maintain the flow of wizard. As this is specially available for GCC, this will remain hidden for newcomers.

You'll require just One additional click to activate this feature.

Please post your valuable opinion. :D

I've not committed the code to svn yet.

TDragon:
In my valuable opinion ( 8) ) that should work perfectly. (Although I could get extremely nitpicky and say that, according to human interface guidelines, checkboxes should be checked to indicate affirmation and unchecked to indicate negation. But I could hardly care less.)

At any rate, I'd like to know whenever you commit so I can give it another run-through with GCC and VC2005; I haven't actually tried the wizard in quite a while.

Biplab:

--- Quote from: TDragon on February 15, 2007, 05:23:59 pm ---In my valuable opinion ( 8) ) that should work perfectly.

--- End quote ---

Thanks for ur valuable opinion. :)


--- Quote from: TDragon on February 15, 2007, 05:23:59 pm ---(Although I could get extremely nitpicky and say that, according to human interface guidelines, checkboxes should be checked to indicate affirmation and unchecked to indicate negation. But I could hardly care less.)

--- End quote ---

Thanks for the point. I'd change it to Use __WXDEBUG__ and Debug wxWidgets lib and keep it unchecked. I hope that would be acceptable. ;)


--- Quote from: TDragon on February 15, 2007, 05:23:59 pm ---At any rate, I'd like to know whenever you commit so I can give it another run-through with GCC and VC2005; I haven't actually tried the wizard in quite a while.

--- End quote ---

I'd PM you and Pecan after I commit. :)

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