如何用Python CFFI正确包装C库

Question

我试图包装一个非常简单的C库，只包含两个。C源文件:dbc2dbf.c和blast.c

我正在做以下事情(来自文档):

import os
from cffi import FFI
blastbuilder = FFI()
ffibuilder = FFI()
with open(os.path.join(os.path.dirname(__file__), "c-src/blast.c")) as f:
    blastbuilder.set_source("blast", f.read(), libraries=["c"])
with open(os.path.join(os.path.dirname(__file__), "c-src/blast.h")) as f:
    blastbuilder.cdef(f.read())
blastbuilder.compile(verbose=True)
with open('c-src/dbc2dbf.c','r') as f:
    ffibuilder.set_source("_readdbc",
                          f.read(),
                          libraries=["c"])
with open(os.path.join(os.path.dirname(__file__), "c-src/blast.h")) as f:
    ffibuilder.cdef(f.read(), override=True)
if __name__ == "__main__":
    # ffibuilder.include(blastbuilder)
    ffibuilder.compile(verbose=True)

这不是很有效。我想我没有包括blast.c正确;

有人能帮忙吗?

Answer 1

解决方案(经过测试):

import os
from cffi import FFI
ffibuilder = FFI()
PATH = os.path.dirname(__file__)
with open(os.path.join(PATH, 'c-src/dbc2dbf.c'),'r') as f:
    ffibuilder.set_source("_readdbc",
                          f.read(),
                          libraries=["c"],
                          sources=[os.path.join(PATH, "c-src/blast.c")],
                          include_dirs=[os.path.join(PATH, "c-src/")]
                          )
ffibuilder.cdef(
    """
    static unsigned inf(void *how, unsigned char **buf);
    static int outf(void *how, unsigned char *buf, unsigned len);
    void dbc2dbf(char** input_file, char** output_file);
    """
)
with open(os.path.join(PATH, "c-src/blast.h")) as f:
    ffibuilder.cdef(f.read(), override=True)
if __name__ == "__main__":
    ffibuilder.compile(verbose=True)