...与构造函数不允许在联合问题 [英] ... with constructor not allowed in union problem
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问题描述
我迫切需要为以下问题找到解决方案:
I desperately need to find a solution for the following problem:
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,但是f& ff成员无法访问:
It's frustrating that if I add tags to p1 and p2 structs, the code will compile, but the f & ff members would not be accessible:
...
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'
}
是否有任何方式使这项工作以某种方式?
Is there any way to make this work somehow?
推荐答案
如@Als所说,union不能定义非POD作为成员数据,有一个替代方法。您还可以定义指向非POD的指针作为联合的成员数据。
As @Als said, union cannot define non-POD as member data, there is one alternative. You can still define a pointer to the non-POD as member data of the union.
因此允许这样:
union
{
struct
{
Flags<4096> *pf; //pointer to non-POD
}p1;
struct
{
Flags<16384> *pff; //pointer to non-POD
}p2;
}parts;
但是 Boost.Variant 是更好的选择。
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