cpython_embed.md

Embed CPython Script

The script below is marked with @embed which turns it into a static string in the final C++, and is run after cpython.initalize() is called.

@embed

class A():
	def __init__(self):
		self.value = 100
	def pymethod(self):
		print 'from cpython: self.value=', self.value
		if hasattr(self, 'my_dynamic_var'):
			print self.my_dynamic_var

	def add(self, a,b):
		return a+b

def foo():
	print 'foo'
	return A()

CPython CAPI

The code below shows how to use the cpython module that wraps around the CPython C-API. https://docs.python.org/2/c-api/object.html

Below a.value as int it is trival to know that a is a PyObject and that using the . operator on it makes Rusthon generate the code that calls the CPython C-API. Translation to C++:

auto v = static_cast<int>(PyInt_AS_LONG(PyObject_GetAttrString(a,"value")));

The syntax is used -> to make it explicit that the object is a PyObject, and to make Rusthon generate the required CPython C-API calls.
For example b->pymethod() becomes this C++:

PyObject_Call(
	PyObject_GetAttrString(b,"pymethod"),
	Py_BuildValue("()"),
	NULL
);

Build Options

• @link:python2.7
• @include:/usr/include/python2.7

#backend:c++
import cpython

def main():
	state = cpython.initalize()
	with gil:
		a = cpython.foo()
		print 'addr of a:', a
		print a->value as int
		a->pymethod()

		b = a
		b->pymethod()
		print b->value as int
		v = b->value as int
		print v + 400
		c = b->pymethod()

		r = b->add(1, 2) as int
		print r
		u = b->add(
			b->value, 
			b->value 
		) as int
		print u

		if hasattr(b, "value"):
			print('builtin `hasattr` works on b')
			print str(b)
			b->value = 'set from c++' as pystring
			b->pymethod()
			b->my_dynamic_var = 'hello dynamic var' as pystring
			a->pymethod()
			setattr(a, 'my_dynamic_var', 'setattr OK' as pystring)
			a->pymethod()

			s = b->my_dynamic_var as string
			print s

		pyob = cpython.foo()
		b->my_dynamic_var = pyob
		b->pymethod()

	cpython.finalize(state)