Question

我有C的密码:

typedef result function_callback(struct mes_t* message, void* data) 
struct mes_t
{
uint32_t field1
uint32_t field2
void* data
};
function_one(&function_callback, data)

The application calls the user-defined (in the function_one) callback function function_callback. In the callback function passed field1, field2 and data parameters (data is usually equal to 0)

此示例中 python 的代码是否正确?

class mes_t(ctypes.Structure):
    pass
mes_t._fields_ = [
    ( field1 , ctypes.c_uint32),
    ( dfield2 , ctypes.c_uint32),
    ( data , ctypes.POINTER(ctypes.c_void_p))]
data_t=ctypes.c_void_p
data=data_t()
CALLBACK=CFUNCTYPE(ccg_msg, data_t)
cb_func=CALLBACK()
result = function_one(ctypes.byref(cb_func), ctypes.byref(data))

Answer 1

我在这里猜到了解释你的代码的正确方法。

typedef int /* or whatever */ result;

struct mes_t
{
    uint32_t field1;
    uint32_t field2;
    void* data;
};
typedef result function_callback(struct mes_t* message, void* data);
result function_one(function_callback fcb, void* data);

使用 force_ one_ one () 的 ctypes Python 示例如下:

class mes_t(ctypes.Structure):
    _fields_ = (
        ( field1 , ctypes.c_uint32),
        ( field2 , ctypes.c_uint32),
        ( data , ctypes.c_void_p))

result_t = ctypes.c_int; # or whatever

callback_type = ctypes.CFUNCTYPE(result_t, ctypes.POINTER(mes_t), ctypes.c_void_p)
function_one.argtypes = (callback_type, ctypes.c_void_p)
function_one.restype = result_t

data_p = ctypes.c_char_p( whatever )

def the_callback(mes_p, data_p):
    my_mes = mes_p[0]
    my_data_p = ctypes.cast(data_p, ctypes.c_char_p)  # or whatever
    my_data = my_data_p.value
    print "I got a mes_t object! mes.field1=%r, mes.field2=%r, mes.data=%r, data=%r" 
          % (my_mes.field1, my_mes.field2, my_mes.data, my_data)
    return my_mes.field1

result = function_one(callback_type(the_callback), ctypes.cast(data_p, ctypes.c_void_p))

你可以看到,这个代码和你的代码之间有许多不同之处;可能太多无法对一切做出充分解释。但是,如果有些内容似乎特别令人困惑,我可以解释几个特定部分。总的来说,必须很好地了解Ctype 指针是如何工作的(例如,你可能不想让指针指针无效,但这正是你的 Python 代码所做的 ) 。

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