I've got exported class A and I'd like to avoid reporting private methods when
users type help (mymodule.A). Is there anything like __all__ for modules? naming
these methods starting with _ does not help either.
Please advise,
Gennadiy
___
Cplusplus-si
Jim Bosch gmail.com> writes:
>
> On Fri, 2010-03-19 at 19:55 +, Gennadiy Rozental wrote:
> > I am doing something like this in export part of mymodule:
> >
> > scope S = class_("X", no_init );
> >
> > class_( "Y",... )
> > ...
> > ;
> >
> > In Python I do see mymodule.X.Y, but mymodule.X.
On Fri, 2010-03-19 at 19:55 +, Gennadiy Rozental wrote:
> I am doing something like this in export part of mymodule:
>
> scope S = class_("X", no_init );
>
> class_( "Y",... )
> ...
> ;
>
> In Python I do see mymodule.X.Y, but mymodule.X.Y.__name__ is 'mymodule.Y'
>
Hmm. In the same situa
I am doing something like this in export part of mymodule:
scope S = class_("X", no_init );
class_( "Y",... )
...
;
In Python I do see mymodule.X.Y, but mymodule.X.Y.__name__ is 'mymodule.Y'
Gennadiy
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Cplusplus-sig@pytho
How do I provide doc string to the property like this:
...
.add_property( "goo", make_function(...) )
While we are at it, how do I provide docstring and arg names to the operators:
...
.def( self < self )
Gennadiy
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Cplusplus-sig mailing lis
Hi all,
I want to create python module for class that has protected operator=.
That class placed in std::vector and code generated by Py++ uses that
operator.
Simple example...
*
The c++ code:
#include
class JobPropertyWrapper {
protected:
JobPropertyWrapper& ope
