我迫切需要找到解决以下问题的方法：
namespace test
{
template <int param = 0> struct Flags
{
int _flags;
Flags()
{
_flags = 0;
}
Flags(int flags)
{
_flags = flags;
}
void init()
{
}
};
union example
{
struct
{
union
{
struct
{
Flags<4096> f;
}p1; //error: member 'test::example::<anonymous struct>::<anonymous union>::<anonymous struct> test::example::<anonymous struct>::<anonymous union>::p1' with constructor not allowed in union
struct
{
Flags<16384> ff;
}p2; //error: member 'test::example::<anonymous struct>::<anonymous union>::<anonymous struct> test::example::<anonymous struct>::<anonymous union>::p2' with constructor not allowed in union
}parts;
byte bytes[8];
}data;
int data1;
int data2;
}
}
令人沮丧的是,如果我将标签添加到p1和p2结构,代码将编译,但f& ff成员无法访问：
...
struct p1
{
Flags<4096> f;
};
struct p2
{
Flags<4096> ff;
};
...
void test()
{
example ex;
ex.data.bytes[0] = 0; //Ok
ex.data.parts.p1.f.init(); //error: invalid use of 'struct test::example::<anonymous struct>::<anonymous union>::p1'
}
有什么办法让这项工作以某种方式？
解决方法
正如@Als所说,联盟不能将非POD定义为成员数据,还有一种选择.您仍然可以将指向非POD的指针定义为union的成员数据.
所以这是允许的：
union
{
struct
{
Flags<4096> *pf; //pointer to non-POD
}p1;
struct
{
Flags<16384> *pff; //pointer to non-POD
}p2;
}parts;
但是Boost.Variant是一个更好的选择.
