On 12/26/14 2:12 AM, ilias wrote:
It's not a bug, it's as designed. Boost.python tries function overloads
in reverse registration order and picks the first one that works, in the
sense that all the arguments convert.
Is that behavior specified somewhere in the documentation?
I'm not sure it
Alex,
> It's not a bug, it's as designed. Boost.python tries function overloads
> in reverse registration order and picks the first one that works, in the
> sense that all the arguments convert.
Is that behavior specified somewhere in the documentation?
__
On 12/19/14 5:34 AM, Stefan Seefeld wrote:
But if I declare the function f in reversed order:
BOOST_PYTHON_MODULE(test)
{
def("f", f_double);
def("f", f_int);
}
I will see "invoked f(int n = 5)" when I call f(5) and "invoked f(double d =
3.14)" when I call f(3.14) as it has to be. Why doe
On 18/12/14 06:13 AM, ilias wrote:
> I got an overloaded function:
>
> void f(int n)
> {
> std::clog << "invoked f(int n = " << n << ")" << std::endl;
> }
>
> void f(double d)
> {
> std::clog << "invoked f(double d = " << d << ")" << std::endl;
> }
>
> If I put declarations in that order:
>
> v
