Here is my example:
---
CPP Code:
---
#include
using namespace boost::python;
void updateDoubleReference(double& x)
{
Here is an example:
// c++ code
class A
{
public:
enum ATypes
{
ONE,
TWO,
THREE
};
};
// boost python code
object obj_A =
class_
("A", "A simple A object", init<>() )
;
{
scope in_A(obj_A);
enum_
("ATypes")
.value("ONE", A::ONE)
What happens if you do:
print canupo.read_msc_header(mscfile, scales, float(ptnparams))
?
From: Cplusplus-sig
[mailto:cplusplus-sig-bounces+herron=ellington@python.org] On Behalf Of
Zhan Li
Sent: Tuesday, November 03, 2015 5:13 PM
To: cplusplus-sig@python.org
Subject: [C++-sig] Expose a C+
For the following code:
---
#include
using namespace boost::python;
namespace // unnamed
{
class NullableDouble
{
public:
Nullabl
You should open a cmd shell tool that comes with VS :
e.g. "VS2010 x86 Native Tools Command Prompt"
or "VS2010 x64 Native Tools Command Prompt"
And then in this shell, run your:
b2 toolset=msvc-10.0 --with-python link=shared
It looks like your python version is 64-bit python (Is that
X86'
So it seems to be picking up python34.lib in preference to python27.lib
Do you think this just depends on the path order?
David
From: Cplusplus-sig
[mailto:cplusplus-sig-bounces+david.aldrich=emea.nec@python.org] On Behalf
Of Liam Herron
Sent: 09 March 2016 14:40
To: Devel
: Re: [C++-sig] Trouble building boost with Visual Studio 2010
Ok, thanks. The console defaults to 3.5, which is the problem.
I'll see what I can do.
From: Cplusplus-sig
[mailto:cplusplus-sig-bounces+david.aldrich=emea.nec@python.org] On Behalf
Of Liam Herron
Sent: 09 March 2016 16:
please why it's still choosing Python 3.4 instead of 2.7 please?
Best regards
David
From: Cplusplus-sig
[mailto:cplusplus-sig-bounces+david.aldrich=emea.nec....@python.org] On Behalf
Of Liam Herron
Sent: 09 March 2016 17:03
To: Development of Python/C++ integration
Subject: Re: [C++-sig]
I followed the test examples that I found on the web, and I am able to
replicate the examples.
My issue occurs when I:
1) Write a C++ base class
2) Wrap this C++ base class in C++
3) Inherit from this wrapped C++ class in python
4) Then try to use this python inherited cla
So I have a boost python extension C++ library. I would like to access the
"python commandline" in C++ without having to pass in the "sys.argv" into my
C++ extension.
Is there a way to do this? Include some boost python header and read it from
some init variables or something?
e.g.
python s
trying to avoid additional python wrappers around my boost python library.
Can this be done without additional python? What is the best/cleanest way to
do this?
Regards,
Liam Herron
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