I know that a Numba-jitted function calling another jitted function will recognize this and automatically use a fast C calling convention rather than going through the Python object layer, and therefore avoid the high Python function call overhead:
import numba
@numba.jit
def foo(x):
return x**2
@numba.jit
def bar(x):
return 4 * foo(x) # this will be a fast function call
My question is whether the same is true if I call a Cython function from Numba. So let's say I have a Cython module, foo.pyx:
cpdef double foo(double x):
return x**2
As well as a standard Python module bar.py:
import numba
import foo
@numba.jit
def bar(x):
return 4 * foo.foo(x) # will this be a fast function call?
Will Numba recognize foo.foo as a C-callable function automatically or do I need to tell it manually by, say, setting up a CFFI wrapper?
EDIT: Upon further reflection, Cython functions are just standard "builtin" functions from the view of the Python interpreter. So the question can be made more general: does Numba optimize calls to builtin functions and methods to bypass the Python calling overhead?
There is a limited set of builtin functions (from both the python standard library and numpy) that numba knows how to translate into native code:
Anything else will not be able to be jitted by Numba in nopython mode, thus resorting to objectmode which is much slower.
There is no direct way to pass a cython function to Numba and have it be recognized in nopython mode. Numba does have hooks for cffi:
http://numba.pydata.org/numba-doc/latest/reference/pysupported.html#cffi
that can be leveraged to call outside C code, that you might be able to rig up to call cython if you could create a low level wrapper at the C-level; I'm not 100% sure if this is possible though. I wrote about doing this for calling RMath functions from Numba:
It might be helpful in getting you started if you go that route.