Question

我对ctypes和C都是新手，无法将struct指针变量传递给通过ctypes在python中调用的c函数。如果它太基础和太明显，请忍受。以下是我的C代码的外观。

#include "mylib.h"   (inside this mylib.h file MYSTRUCT is defined)

struct MYSTRUCT* modifystruct(a,b,c,d,e)
{
MYSTRUCT *mystpointer;
.....
.....
return mystpointer;
} 


int mycfunction(mystpointer)
MYSTRUCT *mystpointer;
{
.........
.........
.........
}

像上面一样，modifystruct函数update * mystpointer是指向MYSTRUCT的指针并返回它。 mycfunction是传递返回的mystpointer。在C中，这在主函数内部工作正常。但是，当我尝试使用ctypes将“ .so”文件加载到python中时，它失败了，并且我认为我没有为mystpointer正确定义argtype。以下是我编写的简短python代码。假设上面的c代码被编译为“ mycmodule.so”。

mylib=cdll.LoadLibrary("mycmodule.so")
mycfunction=mylib.mycfunction
mycfunction.restype=c_int
mycfunction.argtypes=[c_void_p]
mystpointer=c_void_p()

在C代码中，我必须将mystpointer类型定义为“ MYSTRUCT * mystpointer;”。但是，我不知道如何在ctypes中执行此操作……相反，我将类型定义为c_void_p，但这会触发失败。预先感谢！

Answer 1

我认为您可能缺少的是确切知道要在哪里分配结构内存。下面的c代码提供了一个为该结构分配内存并返回指向它的指针（new_struct（））的函数。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int a;
    int b;
} my_struct;

my_struct *new_struct()
{
    my_struct *struct_instance = (my_struct *)malloc(sizeof(my_struct));
    memset(struct_instance, 0, sizeof(my_struct));
    return struct_instance;
}

int modify_struct(my_struct *ms) {
    ms->a = 1;
    ms->b = 2;
    return 0;
}

void print_struct_c(my_struct *ms) {
    printf("my_struct {\n"
           "    a = %d\n"
           "    b = %d\n"
           "}\n", ms->a, ms->b);
}

在Python中，要获取指针，请调用进行分配的C函数，然后可以将其传递给其他将其作为参数的C函数。

import ctypes

lib_file_path = <<< path to lib file >>>

# Very simple example of how to declare a ctypes structure to twin the
# C library's declaration. This doesn't need to be declared if the Python
# code isn't going to need access to the struct's data members.
class MyStruct(ctypes.Structure):
    _fields_ = [('a', ctypes.c_int),
                ('b', ctypes.c_int)]

def print_struct(s):
    # Print struct that was allocated via Python ctypes.
    print("my_struct.a = %d, my_struct.b = %d" % (s.a, s.b))

def print_struct_ptr(sptr):
    # Print pointer to struct. Note the data members of the pointer are 
    # accessed via 'contents'.
    print("my_struct_ptr.contents.a = %d, my_struct_ptr.contents.b = %d" 
          % (sptr.contents.a, sptr.contents.b))


my_c_lib = ctypes.cdll.LoadLibrary(lib_file_path)

# If you don't need to access the struct's data members from Python, then 
# it's not necessary to declare MyStruct above. Also, in that case,
# 'restype' and 'argtypes' (below) can be set to ctypes.c_void_p instead.
my_c_lib.new_struct.restype     =  ctypes.POINTER(MyStruct)
my_c_lib.modify_struct.argtypes = [ctypes.POINTER(MyStruct)]

# Call C function to create struct instance.
my_struct_c_ptr = my_c_lib.new_struct()
print_struct_ptr(my_struct_c_ptr)

my_c_lib.modify_struct(my_struct_c_ptr)
print_struct_ptr(my_struct_c_ptr)

# Allocating struct instance from Python, then passing to C function.
my_struct_py = MyStruct(0, 0)
print_struct(my_struct_py)

my_c_lib.modify_struct(ctypes.byref(my_struct_py))
print_struct(my_struct_py)

# Data members of Python allocated struct can be acessed directly.
my_struct_py.a = 555

my_c_lib.print_struct_c(ctypes.byref(my_struct_py)) # Note use of 'byref()'
                                                    # to invoke c function.

上面的代码已更新，包括一个示例，该示例说明如何通过Python分配结构实例，以及如何访问C分配或Python分配结构的数据成员（注意打印功能的差异）。

传递在.h文件中定义的结构指针的Python Ctype。

1 个答案: