linking a Python module: difference between linking with `ld` and with `cc`

Question

This works:

cc leveldb_ext.cc leveldb_object.cc -o leveldb.so -I /usr/include/python2.7 -lpython2.7 -lleveldb -lsnappy -shared -lc

This does not work:

cc -I /usr/include/python2.7 -g -c leveldb_ext.cc leveldb_object.cc
ld -shared -o leveldb.so -lpython2.7 -lleveldb -lsnappy leveldb_ext.o leveldb_object.o -lc

In both cases, I don't get any compiler/linking errors. However, when trying to import it, I get this error:

$ python -c "import leveldb"
Traceback (most recent call last):
  File "<string>", line 1, in <module>
ImportError: ./leveldb.so: undefined symbol: _ZNK7leveldb6Status8ToStringEv

Why? Is there any difference between the two methods? What is the difference?

Answer 1

The order of the object files and the libraries is not the same between the two cases. The order is significant.

Answer 2

Usually a setup.py script is used to compile Python modules. Something like this should work:

from setuptools.extension import Extension

ext_modules = [
    Extension(
        'yourmodule',
        sources=['yourmodule.c'],
        libraries=['a', 'b', 'c'],
        extra_compile_args=['-Wall', '-g'],
    )
]

setup(..., ext_modules=ext_modules)

Setuptools makes sure the compiler and linker will be called with the right flags, avoiding issues like the one in your question.

Btw, you should have a look at Plyvel if you want a nice Python API for LevelDB. See https://github.com/wbolster/plyvel and https://plyvel.readthedocs.org/ for more information. (Disclaimer: I'm the author.)