从 Python 调用 C++ 代码(使用 ctypes)时，如何取回相同的对象(通过引用传递)

CPython中的id()返回Pythonctypes包装对象的地址，而不是包装对象的地址.为此，请使用ctypes.addressof().

ctypes也只理解C普通老式数据(POD)类型和 struct .它不知道C++名称空间或引用，但由于引用实际上是指针的C++语法糖，您可以使用.argtypes和.restype中的指针作为替代.

下面是一个最小的例子:

test.cpp个

#include <stdio.h>

namespace PEParserNamespace {
struct PEParserBase {
    void* hFile;
    unsigned long dwFileSize;
    unsigned long bytes;
    void* fileBuffer;
};
}

extern "C" __declspec(dllexport)
PEParserNamespace::PEParserBase& _cdecl test(PEParserNamespace::PEParserBase* base) {
    printf("the C++ function was called\n");
    base->bytes = 12345;
    return *base;
}

test.py个

import ctypes as ct

class PEParserBase(ct.Structure):
    _fields_ = (("hFile", ct.c_void_p),
                ("dwFileSize", ct.c_ulong),
                ("bytes", ct.c_ulong),
                ("fileBuffer",ct.c_void_p))

dll = ct.CDLL('./test')
dll.test.argtypes = ct.POINTER(PEParserBase),
dll.test.restype = ct.POINTER(PEParserBase)    # Use a pointer here for the reference

base = PEParserBase()
pbase = dll.test(ct.byref(base))

print(hex(ct.addressof(base)), base.bytes)
print(hex(ct.addressof(pbase.contents)), base.bytes)  # .contents dereferences the pointer
                                                  # so we get the address of the structure
                                                  # not the address of the pointer itself.

输出:

the C++ function was called, base=00000169712ADAF0
0x169712adaf0 12345
0x169712adaf0 12345