我在Cython中编写Python 2.7扩展模块。 如何创建实现新型缓冲区接口的Python对象，该接口包装由C库提供给我的内存块？
内存块只是一串字节，而不是结构或多维数组。我得到了一个const void *指针和一个长度，以及一些有关指针保持有效时间的详细信息。

我无法复制内存，这会降低应用程序的性能。

使用旧样式的缓冲区对象，我可以简单地使用PyBuffer_FromMemory()，但是我似乎找不到找到类似的简单方法来生成新样式的缓冲区对象。

我是否必须创建自己的实现缓冲区接口的类？还是Cython提供了一种简便的方法？

我已经从Cython文档中阅读了Unicode和Passing
Strings
和Typed
Memoryviews页面，但是该文档不够精确，也不是很完整，并且没有类似于我想要的示例的示例。

这是我尝试过的（test.pyx）：

from libc.stdlib cimport malloc
from libc.string cimport memcpy

## pretend that this function is in some C library and that it does
## something interesting.  (this function is unrelated to the problem
## I'm experiencing -- this is just an example function that returns a
## chunk of memory that I want to wrap in an object that follows the
## new buffer protocol.)
cdef void dummy_function(const void **p, size_t *l):
    cdef void *tmp = malloc(17)
    memcpy(tmp, "some test\0 bytes", 17)
    p[0] = tmp
    l[0] = 17

cpdef getbuf():
    cdef const void *cstr
    cdef size_t l
    dummy_function(&cstr, &l)

    ## error: test.pyx:21:20: Invalid base type for memoryview slice: void
    #cdef const void[:] ret = cstr[:l]

    ## error: test.pyx:24:9: Assignment to const 'ret'
    #cdef const char[:] ret = cstr[:l]

    ## error: test.pyx:27:27: Cannot convert 'void const *' to memoryviewslice
    #cdef char[:] ret = cstr[:l]

    ## this next attempt cythonizes, but raises an exception:
    ## $ python -c 'import test; test.getbuf()'
    ## Traceback (most recent call last):
    ##   File "<string>", line 1, in <module>
    ##   File "test.pyx", line 15, in test.getbuf (test.c:1411)
    ##   File "test.pyx", line 38, in test.getbuf (test.c:1350)
    ##   File "stringsource", line 614, in View.MemoryView.memoryview_cwrapper (test.c:6763)
    ##   File "stringsource", line 321, in View.MemoryView.memoryview.__cinit__ (test.c:3309)
    ## BufferError: Object is not writable.
    cdef char[:] ret = (<const char *>cstr)[:l]

    ## this raises the same exception as above
    #cdef char[:] ret = (<char *>cstr)[:l]

    return ret