我正在处理C中的联合,我希望有一个函数模板,它可以根据模板参数访问活动的union成员.
代码就像(doSomething只是一个例子)：
union Union {
int16_t i16;
int32_t i32;
};
enum class ActiveMember {
I16,I32
}
template <ActiveMember M>
void doSomething(Union a,const Union b) {
selectMemeber(a,M) = selectMember(b,M);
// this would be exactly (not equivalent) the same
// that a.X = b.X depending on T.
}
为了实现这一点,我只发现了一些糟糕的黑客攻击,比如专业化,或者是一种不同类的访问和分配方式.
我错过了什么,这样的事情应该用其他方法来做？
解决方法
可能性1
而不是使用枚举,您可以使用简单的结构来选择成员：
typedef short int16_t;
typedef long int32_t;
union Union {
int16_t i16;
int32_t i32;
};
struct ActiveMemberI16 {};
struct ActiveMemberI32 {};
template <typename M>
void doSomething(Union& a,Union b) {
selectMember(a,M()) = selectMember(b,M());
// this would be exactly (not equivalent) the same
// that a.X = b.X depending on T.
}
int16_t& selectMember(Union& u,ActiveMemberI16)
{
return u.i16;
}
int32_t& selectMember(Union& u,ActiveMemberI32)
{
return u.i32;
}
int main(int argc,char* argv[])
{
Union a,b;
a.i16 = 0;
b.i16 = 1;
doSomething<ActiveMemberI16>(a,b);
std::cout << a.i16 << std::endl;
b.i32 = 3;
doSomething<ActiveMemberI32>(a,b);
std::cout << a.i32 << std::endl;
return 0;
}
这需要为union中的每个成员定义struct和selectMember方法,但至少可以在许多其他函数中使用selectMember.
请注意,我将参数转换为引用,如果不合适,您可以调整它.
可能性2
通过将union指针强制转换为所需的类型指针,您可以使用单个selectMember函数.
typedef short int16_t;
typedef long int32_t;
union Union {
int16_t i16;
int32_t i32;
};
template <typename T>
T& selectMember(Union& u)
{
return *((T*)&u);
}
template <typename M>
void doSomething(Union& a,Union b) {
selectMember<M>(a) = selectMember<M>(b);
// this would be exactly (not equivalent) the same
// that a.X = b.X depending on T.
}
int _tmain(int argc,_TCHAR* argv[])
{
Union a,b;
a.i16 = 0;
b.i16 = 1;
doSomething<int16_t>(a,b);
std::cout << a.i16 << std::endl;
b.i32 = 100000;
doSomething<int32_t>(a,b);
std::cout << a.i32 << std::endl;
return 0;
}
