Get the name of the extension file built by CFFI
If one builds a CFFI-based extension as
ffi = FFI()
ffi.cdef("...code...")
ffi.set_source("_example", "...code...")
ffi.compile()
a compiled extension file is created, for example on my system it's "_example.cpython-36m-darwin.so".
It can be imported in a semi-hacky way without knowing the full file name as
sys.path.append(path)
from _example import lib
sys.path.pop()
This is almost safe, because an FFI extension does not modify sys.path, and it is very unlikely to do so in the future. Nevertheless, it would be nice to do a fully stateless import, via importlib:
spec = importlib.util.spec_from_file_location(
"_example", path + "/_example.cpython-36m-darwin.so")
mod = importlib.util.module_from_spec(spec)
spec.loader.exec_module(mod)
lib = mod.lib
The problem is that it requires the full path to the extension file. Is there some way to get this full name, with the OS-dependent suffix and extension, knowing only the module name ("_example"), either through cffi or standard library means?
python python-cffi
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
add a comment |
If one builds a CFFI-based extension as
ffi = FFI()
ffi.cdef("...code...")
ffi.set_source("_example", "...code...")
ffi.compile()
a compiled extension file is created, for example on my system it's "_example.cpython-36m-darwin.so".
It can be imported in a semi-hacky way without knowing the full file name as
sys.path.append(path)
from _example import lib
sys.path.pop()
This is almost safe, because an FFI extension does not modify sys.path, and it is very unlikely to do so in the future. Nevertheless, it would be nice to do a fully stateless import, via importlib:
spec = importlib.util.spec_from_file_location(
"_example", path + "/_example.cpython-36m-darwin.so")
mod = importlib.util.module_from_spec(spec)
spec.loader.exec_module(mod)
lib = mod.lib
The problem is that it requires the full path to the extension file. Is there some way to get this full name, with the OS-dependent suffix and extension, knowing only the module name ("_example"), either through cffi or standard library means?
python python-cffi
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
add a comment |
If one builds a CFFI-based extension as
ffi = FFI()
ffi.cdef("...code...")
ffi.set_source("_example", "...code...")
ffi.compile()
a compiled extension file is created, for example on my system it's "_example.cpython-36m-darwin.so".
It can be imported in a semi-hacky way without knowing the full file name as
sys.path.append(path)
from _example import lib
sys.path.pop()
This is almost safe, because an FFI extension does not modify sys.path, and it is very unlikely to do so in the future. Nevertheless, it would be nice to do a fully stateless import, via importlib:
spec = importlib.util.spec_from_file_location(
"_example", path + "/_example.cpython-36m-darwin.so")
mod = importlib.util.module_from_spec(spec)
spec.loader.exec_module(mod)
lib = mod.lib
The problem is that it requires the full path to the extension file. Is there some way to get this full name, with the OS-dependent suffix and extension, knowing only the module name ("_example"), either through cffi or standard library means?
python python-cffi
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
If one builds a CFFI-based extension as
ffi = FFI()
ffi.cdef("...code...")
ffi.set_source("_example", "...code...")
ffi.compile()
a compiled extension file is created, for example on my system it's "_example.cpython-36m-darwin.so".
It can be imported in a semi-hacky way without knowing the full file name as
sys.path.append(path)
from _example import lib
sys.path.pop()
This is almost safe, because an FFI extension does not modify sys.path, and it is very unlikely to do so in the future. Nevertheless, it would be nice to do a fully stateless import, via importlib:
spec = importlib.util.spec_from_file_location(
"_example", path + "/_example.cpython-36m-darwin.so")
mod = importlib.util.module_from_spec(spec)
spec.loader.exec_module(mod)
lib = mod.lib
The problem is that it requires the full path to the extension file. Is there some way to get this full name, with the OS-dependent suffix and extension, knowing only the module name ("_example"), either through cffi or standard library means?
python python-cffi
python python-cffi
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
asked Nov 16 '18 at 2:26
fjarrifjarri
7,5922840
7,5922840
add a comment |
add a comment |
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