如何使用 ctypes 将 Python 列表转换为 C 数组？
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How do I convert a Python list into a C array by using ctypes?
提问by No One in Particular
If I have the follow 2 sets of code, how do I glue them together?
如果我有以下两套代码，我该如何将它们粘在一起？
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.
如何使用 x 中的连续元素列表调用 c_function？我试图将 x 转换为 c_void_p，但这没有用。
I also tried to use something like
我也尝试使用类似的东西
x = c_void_p * 10
for e in y:
x[i] = e
but this gets a syntax error.
但这会出现语法错误。
The C code clearly wants the address of an array. How do I get this to happen?
C 代码显然需要数组的地址。我怎样才能做到这一点？
采纳答案by Gabi Purcaru
The following code works on arbitrary lists:
以下代码适用于任意列表：
import ctypes
pyarr = [1, 2, 3, 4]
arr = (ctypes.c_int * len(pyarr))(*pyarr)
回答by Ryan Ginstrom
回答by akhan
This is an explanation of the accepted answer.
这是对已接受答案的解释。
ctypes.c_int * len(pyarr)creates an array (sequence) of type c_intof length 4 (python3, python 2). Since c_intis an object whose constructor takes one argument, (ctypes.c_int * len(pyarr)(*pyarr)does a one shot init of each c_intinstance from pyarr. An easier to readform is:
ctypes.c_int * len(pyarr)创建一个c_int长度为 4（python3，python 2）类型的数组（序列）。由于c_int是一个对象，其构造函数接受一个参数，因此(ctypes.c_int * len(pyarr)(*pyarr)对每个c_int实例进行一次一次初始化pyarr。一个更容易阅读的表格是：
pyarr = [1, 2, 3, 4]
seq = ctypes.c_int * len(pyarr)
arr = seq(*pyarr)
Use typefunction to see the difference between seqand arr.
使用type功能看之间的区别seq和arr。
回答by J?rg Beyer
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:
这将给出：
1
2
3
