在 2020/2/4 上午2:37, Edward Diener 写道:
> On 2/2/2020 8:11 AM, Liu Hao wrote:
>> 在 2020/2/1 下午7:06, Edward Diener 写道:
>>> Given the code:
>>>
>>> class cbase;
>>> int main()
>>>     {
>>>     typedef int __attribute__ ((__stdcall__)) (cbase::* atype)();
>>>     typedef int __attribute__ ((__cdecl__)) (cbase::* btype)();
>>>     typedef int __attribute__ ((__fastcall__)) (cbase::* ctype)();
>>>     typedef int __attribute__ ((__thiscall__)) (cbase::* dtype)();
>>>     return 0;
>>>     }
>>>

> This does not make sense to me as the compiler seems to be objecting to
> just having different pointer to member function types with different
> calling conventions for the same class, as if all member functions for a
> class must have the same calling conventions. Does gcc not allow
> different calling conventions for different member functions of a class,
> and, if so, where is this documented ?
> 
> 

If I substitute the first typedef with an equivalent(?)
using-declaration then I get a warning:

```
test.cc: In function ‘int main()’:
test.cc:5:65: warning: ‘__stdcall__’ attribute only applies to function
types [-Wattributes]
     using atype = int __attribute__ ((__stdcall__)) (cbase::* )();
```

It looks to me that in the typedef case, the stdcall attribute attaches
to the `int`, not the (indirect) function type. So long as GCC parses it
as such, this eliminates the warning:

```
typedef int ( __attribute__ ((__stdcall__ ))  cbase::* atype)();
//          ^_______parenthesis_moved_______/
```

The mangled names of these two types (which can be examined using
`typeid(___).name`) don't differ, so I presume it is kind of a bug in
intermediate representation.


-- 
Best regards,
LH_Mouse