Python ctypes，将c_void_p作为out函数传递给c函数

Question

I'm writing a Python module that wraps some C++ code. I used to use this function to pass a pointer from C++ to Python (the pointer would be freed by a different function

DLLEXPORT void* run(double input1, double input2)

I've added error return codes so my new function looks like

DLLEXPORT int run(double input1, double input2, void* output)

But now I can't seem to get the value of the pointer, in Python I set up the function with ctypes as follows

from ctypes import *
mydll = cdll.mydll

mydll.run.argtypes = [c_double, # input 1
                      c_double, # input 2
                      c_void_p] # output pointer
mydll.run.restype  = c_int   

Then I use the function by creating a new void pointer in Python and passing it to the dll function

p = c_void_p()

retval = mydll.run(1.2, 3.4, p)

print p

After running this code I am left with p equal to c_void_p(None) .

Passing this pointer to other functions causes illegal access exceptions at address 0x0, so I don't think it is being updated.

I am expected some address to populate p after execution. I am missing something with ctypes? I can create a double array by allocating (c_double * 10)() and pass that to c function to be written to, why can't I pass a void pointer for the same purpose?

Answer 1

As eryksun's comment points out, for void* to be an output parameter, it should be void** :

# DLLEXPORT int run(double input1, double input2, void** output)

from ctypes import *
mydll = cdll.mydll

mydll.run.argtypes = [c_double, # input 1
                      c_double, # input 2
                      POINTER(c_void_p)] # output pointer
mydll.run.restype  = c_int

p = c_void_p()
retval = mydll.run(1.2,3.4,byref(p))

print p.contents