如何在Python中获取CFFI函数的`const`修饰符？

Question

I am writing a Python wrapper for generated C code, which contains functions that use pointer arguments as outputs. I want to introspect those functions, generated using CFFI, to determine if they are input our output arguments. The input arguments are marked const , but it seems CFFI discards this information.

Consider the following example

from cffi import FFI
ffibuilder = FFI()

ffibuilder.set_source("_example",
   r"""
        int add(const int a, const int b) {
            return a+b;
        }
    """)

ffibuilder.cdef("""
        int add(const int a, const int b);
""")

if __name__ == "__main__":
    ffibuilder.compile(verbose=True)
    import _example
    ff = _example.lib.add
    ff_t = _example.ffi.typeof(ff)
    print(ff_t)

Which prints <ctype 'int(*)(int, int)'> , while the inputs are defined const in the code.

Answer 1

Sorry, not possible. This info is discarded already when translating the parsed code to CFFI's internal ctype.

Look at the pycparser module (which is the one that CFFI uses internally); it transforms C source code into a parse tree. From there you can get all this info again more directly.