Create CUDAMatrix from device pointer about cudamat HOT 5 OPEN

Is there any C API for CUDAMat? I want to create a matrix using CUDAMat in python from an already allocated GPU memory and I can only access the memory by a pointer in c.

CUDAMat is a C shared library accessed from Python via the ctypes module. So the CUDAMatrix is defined in Python and there is no C API to create it. However, if you can get your pointer into Python, you can create a CUDAMatrix accessing this pointer. See this example:
https://github.com/Theano/Theano/blob/7a97cfe/theano/misc/cudamat_utils.py#L67-L86
In that example, it is assumed that the memory belongs to the already existing object x, so the matrix is marked as not owning the data, and the object x is attached as a property to the CUDAMatrix object so Python will not delete x before deleting the CUDAMatrix. If instead, you want the memory to be managed by cudamat (and freed with cublasFree), just set owns_data to True.

Hope this makes things clear -- if you want to make this use case easier, I'd be open for a PR that adds a factory method CUDAMatrix.from_memory or something, taking a device pointer, shape and an optional parent object.

from cudamat.

@f0k Thanks for your answer! It's clear.
But the python module I'm trying to extend uses boost.python to wrap C++ code and doesn't expose the GPU memory pointer to python. So I have to start from C++ and it seems that boost.python doesn't have an elegant way to expose a pointer such as float * to python. So I'm wondering whether can I use cudamat to handle this along with boost python?

from cudamat.

Or what type of object should I expose in python managing the GPU memory pointer?

from cudamat.

So I have to start from C++ and it seems that boost.python doesn't have an elegant way to expose a pointer such as float * to python.

There should be a way. In the worst case, you can expose it as a Python long and then cast it to a pointer within Python using ctypes, as in the example code I linked to.
You can also create the cudamat struct in C/C++, but then you still have to pass it to Python -- this won't make things easier.

from cudamat.

@f0k I will try it. Thanks for your thorough explanation.

from cudamat.