Oded Padon wrote:
There is another reason why I think this has to be possible.
> It is possible using ctypes. Using ctypes with the same C++
> code written above (except for extern "C"), the following python code:
import ctypes;
lib_cpp = ctypes.CDLL('./test_cpp.so')
result = ctypes.c_int(0);
lib_cpp.add(1, 2, ctypes.byref(result));
print result;
prints: "c_long(3)"
Yes. In this case, you allocated and passed a specific ctypes object
guaranteed to be compatible with a C++ int.
I still stand by my earlier point. Python doesn't recognize the concept
of 'const'. All int values are equally immutable. If Boost.Python were
to allow usage like this:
result = int()
cppmodule.add(1, 2, result)
assert result == 3
then what should happen in this innocent-looking case?
result = int()
cppmodule.add(result, 1, 2)
Or consider this. In recent versions of the Python interpreter, if you
assign (say) 5 to a variable and then double it enough times, you will
eventually produce a value larger than a C++ int. Unlike C++, Python
will grow the storage associated with that variable to be able to
represent the value exactly.
What should happen when you try to pass *that* value to a C++ int&
parameter?
If you're happy with the ctypes approach, you could easily introduce a
convention like this:
class RefInt(object):
def __init__(self, value=int()):
self.value = value
def __int__(self):
return self.value
# ...other operations as desired for convenience...
Now, assuming you've coded your C++ add() function to accept a RefInt,
you can write:
result = RefInt(0)
cppmodule.add(1, 2, result)
assert int(result) == 3
which is just as facile as the ctypes code you cited.
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