将cython分配的缓冲区传递给python [英] Pass a cython allocated buffer to python

问题描述

我知道这篇文章关于将Cython malloc的数据传递给Python.但是我想知道是否在返回之前强制转换为memoryview可以避免内存泄漏:

I am aware of this post about passing over Cython malloc'ed data to Python. I however wonder if casting to a memoryview before returning avoids the memory leak:

def someCythonFunction():
    try:
        mPtr = <double *>PyMem_Malloc(N * M * sizeof(double)
        for n in range(N):
            for m in range(M):
                mPtr[m + n*M]= ...
        return <double[:N,:M]> mPtr
    finally:
        print("In cython %d" % <int>mPtr)
        PyMem_Free(mPtr)

然后在Python中:

then in Python:

mView = someCythonFunction()

这是安全正确的吗?

推荐答案

这是不安全且不正确的.当函数结束时，它将释放内存，这意味着您返回的指针立即无效.

This is unsafe and incorrect. It deallocates the memory as the function ends which means the pointer that you return is immediately invalid.

您需要将内存的生存期与Python对象的生存期联系起来(如您所链接的示例中，在析构函数中释放了内存).推荐的最简单方法是使用numpy数组或标准库array数组(或您选择的其他库).

You need to tie the lifetime of the memory to the lifetime of a Python object (as in the example you linked to, the memory is freed in the destructor). The simplest and recommended way of doing this is to use a numpy array or a standard library array array (or other library of your choice).

如果您无法避免使用malloc，一种选择是使用 cython.view.array类.您可以为其选择分配一个回调函数以用于销毁，并且可以愉快地将其分配给memoryviews:

If you can't avoid using malloc one option is to use the cython.view.array class. You can assign it a callback function of your choice to use on destruction and it can be happily assigned to memoryviews:

cdef double[:,:] mview
cdef double* mPtr = <double *>PyMem_Malloc(N * M * sizeof(double))
a = cython.view.array(shape=(N, M), itemsize=sizeof(double), format="d",
                      mode="C", allocate_buffer=False)
a.data = <char *> mPtr
a.callback_free_data = PyMem_Free

mview = a

您可以安全地返回a或mview(如果您返回a则无需费心mview)，它们将在正确的时间正确获得.

You can safely return either a or mview (if you return a then there's no need to bother with mview) and they will get correctly at the right time.

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