Question

Windows平台上有一些遗留的COM代码，想要移植到Linux，但__uuidof（）不能在gcc下使用

对于大多数情况，可以使用以下代码解决它。

#define __uuidof(T)  IID_ ##T

但在某些情况下，它无法正常工作，例如，

template <class T>
HRESULT QueryInterface(IUnknown* p, T** ppv)
{
    return p->QueryInterface(__uuidof(T), (void**)ppv);
}

所以我想实现一个模板函数让它工作，比如，

template <class T>
GUID __uuidof()
{
    return IID_ ##typeof(T);
}

如何实现它并让它与一般解决方案完全一致？无法保证每个接口定义中都存在静态成员_IID。

感谢。

Answer 1

根据帖子How can I define an UUID for a class, and use __uuidof, in the same way for g++ and Visual C++?虽然我无法在该帖子中成功编译代码，但是在遵循其中提到的想法之后，这个想法很好，并且可以编译并运行良好的示例代码用于解决gcc下的__uuid

#include <iostream>
#include <memory.h>
#include <assert.h>

using namespace std;

#define nybble_from_hex(c)      ((c>='0'&&c<='9')?(c-'0'):((c>='a'&&c<='f')?(c-'a' + 10):((c>='A'&&c<='F')?(c-'A' + 10):0)))
#define byte_from_hex(c1, c2)   ((nybble_from_hex(c1)<<4)|nybble_from_hex(c2))

typedef struct _GUID {
    unsigned long  Data1;
    unsigned short Data2;
    unsigned short Data3;
    unsigned char  Data4[ 8 ];
} GUID;

static GUID GUID_NULL = {0, 0, 0, {0, 0, 0, 0, 0, 0, 0 ,0}};

#ifndef _REFGUID_DEFINED
#define _REFGUID_DEFINED
#ifdef __cplusplus
#define REFGUID const GUID &
#else
#define REFGUID const GUID * __MIDL_CONST
#endif
#endif

__inline int IsEqualGUID(REFGUID rguid1, REFGUID rguid2)
{
    return !memcmp(&rguid1, &rguid2, sizeof(GUID));
}

#ifdef __cplusplus
__inline bool operator==(REFGUID guidOne, REFGUID guidOther)
{
    return !!IsEqualGUID(guidOne,guidOther);
}

__inline bool operator!=(REFGUID guidOne, REFGUID guidOther)
{
    return !(guidOne == guidOther);
}
#endif

struct Initializable : public GUID
{
    explicit
        Initializable(char const (&spec)[37])
        : GUID()
    {
        for (int i = 0; i < 8; ++i)
        {
            Data1 = (Data1 << 4) | nybble_from_hex(spec[i]);
        }
        assert(spec[8] == '-');
        for (int i = 9; i < 13; ++i)
        {
            Data2 = (Data2 << 4) | nybble_from_hex(spec[i]);
        }
        assert(spec[13] == '-');
        for (int i = 14; i < 18; ++i)
        {
            Data3 = (Data3 << 4) | nybble_from_hex(spec[i]);
        }
        assert(spec[18] == '-');
        for (int i = 19; i < 23; i += 2)
        {
            Data4[(i - 19) / 2] = (nybble_from_hex(spec[i])<<4) | nybble_from_hex(spec[i+1]);
        }
        assert(spec[23] == '-');
        for (int i = 24; i < 36; i += 2)
        {
            Data4[2 + (i - 24) / 2] = (nybble_from_hex(spec[i]) << 4) | nybble_from_hex(spec[i + 1]);
        }
    }
};

template<class T>
inline
auto __my_uuidof()
-> GUID const &
{
    return GUID_NULL;
}

#define CPPX_GNUC_UUID_FOR( name, spec )            \
template<>                                          \
inline                                              \
auto __my_uuidof<name>()                            \
    -> GUID const&                                  \
{                                                   \
    using ::operator"" _uuid;                       \
    static GUID the_uuid = spec ## _uuid;           \
                                                    \
    return the_uuid;                                \
}                                                   \
                                                    \
template<>                                          \
inline                                              \
auto __my_uuidof<name*>()                           \
    -> GUID const&                                  \
{ return __my_uuidof<name>(); }                     \
                                                    \
static_assert( true, "" )

auto operator"" _uuid(char const* const s, size_t const size) 
-> GUID
{
    return Initializable(reinterpret_cast<char const (&)[37]>(*s));
}

#define CPPX_UUID_FOR    CPPX_GNUC_UUID_FOR

#define __uuid(T)       __my_uuidof<T>()

struct Foo {};
CPPX_UUID_FOR(Foo, "dbe41a75-d5da-402a-aff7-cd347877ec00");

Foo foo;

template <class T>
void QueryInterface(T** p)
{
    if (p == NULL)
        return;

    GUID guid = "dbe41a75-d5da-402a-aff7-cd347877ec00"_uuid;
    if (__uuid(T) == guid)
    {
        *p = &foo;
    }
    else
    {
        *p = NULL;
    }

    return;
}

int main()
{
    Foo* p = NULL;
    QueryInterface(&p);

    cout << "p: " << p << ", &foo: " << &foo << endl;   

    return 0;
}

保存到uuidtest.cpp，我使用g ++ - 7成功编译（其他gcc与c ++ 11/14支持也可以正常工作），运行良好：

g++-7 uuidtest.cpp -o uuidtest
./uuidtest
p: 0x6021c0, &foo: 0x6021c0

Answer 2

也许typeid会起作用

否则，您可以使用reinterpret_cast。

For exemple here we have the class A;

A * ex;
ex = reinterpret_cast<A>(*pv);
if (ex == NULL)
{
  //Do some stuff
  //this means **ex** isn't a A class
}

它适用于从A继承的所有类。

Deduce模板参数类型名称并生成conjuct变量

2 个答案: