If I have the follow 2 sets of code, how do I glue them together?
void
c_function(void *ptr) {
int i;
for (i = 0; i < 10; i++) {
printf("%p", ptr[i]);
}
return;
}
def python_routine(y):
x = []
for e in y:
x.append(e)
How can I call the c_function with a contiguous list of elements in x? I tried to cast x to a c_void_p, but that didn't work.
I also tried to use something like
x开发者_StackOverflow社区 = c_void_p * 10
for e in y:
x[i] = e
but this gets a syntax error.
The C code clearly wants the address of an array. How do I get this to happen?
The following code works on arbitrary lists:
import ctypes
py_values = [1, 2, 3, 4]
arr = (ctypes.c_int * len(py_values))(*py_values)
This is an explanation of the accepted answer.
ctypes.c_int * len(pyarr)
creates an array (sequence) of type c_int
of length 4 (python3, python 2). Since c_int
is an object whose constructor takes one argument, (ctypes.c_int * len(pyarr)(*pyarr)
does a one shot init of each c_int
instance from pyarr
. An easier to read form is:
pyarr = [1, 2, 3, 4]
seq = ctypes.c_int * len(pyarr)
arr = seq(*pyarr)
Use type
function to see the difference between seq
and arr
.
From the ctypes tutorial:
>>> IntArray5 = c_int * 5
>>> ia = IntArray5(5, 1, 7, 33, 99)
import ctypes
import typing
def foo(aqs : typing.List[int]) -> ctypes.Array:
array_type = ctypes.c_int64 * len(aqs)
ans = array_type(*aqs)
return ans
for el in foo([1,2,3]):
print(el)
this will give:
1
2
3
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