How to pass references of declared variables to C function in python using ctypes

Question

I would like to call a C function from python. This C function is void, thus the "return parameters" (data I want to change) are defined as pointers in the C function's definition.

The function in C looks like this (note: I cannot and will not change it as it is generated automatically by MATLAB codegen):

void func(int *input_var, // input
 float *output_var //output
 ) {
 ... 
}

from python, I am calling my function as so

import ctypes as C
func = C.CDLL('path/to/lib.so').func
input_var = C.c_int(5)
output_var = C.POINTER(C.c_float) # basically want to just declare a pointer here
func(C.byref(input_var), C.byref(output_var))

The error I get is

TypeError: byref() argument must be a ctypes instance, not '_ctypes.PyCPointerType'

if I remove bref() I get

ctypes.ArgumentError: argument 2: <class 'TypeError'>: Don't know how to convert parameter 2

I also tried to pass in output_var as C.byref(output_var()); this leads to a Segmentation fault (core dumped)

• Mark Tolonen: POINTER(c_float) is a type. An instance would be POINTER(c_float)() but what you actually need is just a float instance: float(). Pass it byref
• Brian Barry: creating the instance works.

Answer 1

Listing [Python.Docs]: ctypes - A foreign function library for Python.

Do things like you'd do it from C (and you're already doing for the input argument): declare a float variable and pass its address to the function (via byref).

One important thing: check [SO]: C function called from Python via ctypes returns incorrect value (@CristiFati's answer) for a common pitfall when working with CTypes (calling functions).

Example (simplest scenario: pointers wrapping single values, if there were arrays, or function performed some memory allocations, things would be a bit more complex).

• dll00.c:

  #include <stdio.h>
  #if defined(_WIN32)
  # define DLL00_EXPORT_API __declspec(dllexport)
  #else
  # define DLL00_EXPORT_API
  #endif
  #if defined(__cplusplus)
  extern "C" {
  #endif
  DLL00_EXPORT_API void dll00Func00(int *pIn, float *pOut);
  #if defined(__cplusplus)
  }
  #endif
  void dll00Func00(int *pIn, float *pOut)
  {
   if (pIn)
   printf("C - In: %d\n", *pIn);
   if (pOut) {
   float f = 3.141593 * (pIn ? *pIn : 1);
   printf("C - Setting out to: %.3f\n", f);
   *pOut = f;
   }
  }
  

• code00.py:

  #!/usr/bin/env python
  import ctypes as cts
  import sys
  DLL_NAME = "./dll00.{:s}".format("dll" if sys.platform[:3].lower() == "win" else "so")
  def main(*argv):
   dll = cts.CDLL(DLL_NAME)
   func = dll.dll00Func00
   func.argtypes = (cts.POINTER(cts.c_int), cts.POINTER(cts.c_float))
   func.restype = None
   i = cts.c_int(2)
   f = cts.c_float(0)
   res = func(cts.byref(i), cts.byref(f))
   print("Values after calling function: {:d}, {:.3f}".format(i.value, f.value))
  if __name__ == "__main__":
   print("Python {:s} {:03d}bit on {:s}\n".format(" ".join(elem.strip() for elem in sys.version.split("\n")),
   64 if sys.maxsize > 0x100000000 else 32, sys.platform))
   rc = main(*sys.argv[1:])
   print("\nDone.\n")
   sys.exit(rc)
  

Output:

(qaic-env) [cfati@cfati-5510-0:/mnt/e/Work/Dev/StackOverflow/q075393602]> ~/sopr.sh
### Set shorter prompt to better fit when pasted in StackOverflow (or other) pages ###
[064bit prompt]> ls
code00.py dll00.c
[064bit prompt]>
[064bit prompt]> gcc -shared -o dll00.so dll00.c
[064bit prompt]>
[064bit prompt]> python ./code00.py
Python 3.8.10 (default, Nov 14 2022, 12:59:47) [GCC 9.4.0] 064bit on linux
C - In: 2
C - Setting out to: 6.283
Values after calling function: 2, 6.283
Done.

• Brian Barry: This makes sense, however reading your specific scenario was a little confusing wrt my situation since you are initializing your output variable f with a value (0). But in my case it turns out that with ctypes you can "declare" a variable by instantiating one without an argument/value, i.e., c_float() as explained by @Mark Tolonen
• CristiFati: :) It's like in C: you can have float f; or float f = 0;. I always initialize variables (it's a habit of mine not to rely on compiler defaults). Anyway in CTypes the 2 are equivalent, as if you print the "uninitialized" one it's still 0 (initialized by default). Also it's a good thing in your situation, as you can clearly see that the function actually changed the value.
• Brian Barry: I see; I was unaware of this compiler default behavior!
• CristiFati: Actually I misread the comment. POINTER(c_float)() creates a float* (which is NULL). Now, I don't know what the function does inside (whether it handles that case or not, or if it allocates memory, and so on). Passing a NULL pointer in my function doesn't do anything with it, therefore there's no return value.

Answer 2

you can use the .byref directly after declaring the variable no need to make a pointer.

so in your example, you can declare the output as the variable type you need it to be and then pass that by reference to the function.

 output_var = C.c_float()
 func(C.byref(input_var), C.byref(output_var)

this should work.

• Mark Tolonen: No, it won’t work. c_float is a type. You need an instance: c_float()