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：

instead of using an enum, you can use simple structs to pick the member:

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;
}

这需要为联合中的每个成员定义一个struct和一个selectMember方法，但至少你可以使用selectMember跨越许多其他功能。

This requires to define a struct and a selectMember method for every member in the union, but at least you can use selectMember across many other functions.

请注意，我将参数转为参考，如果不适当，您可以调整此参数。

Note that I turned the arguments to references, you might adjust this if not appropriate.

可能性2

通过将union指针转换为所需的类型指针，您可以使用一个selectMember函数。

By casting the union pointer to the desired type pointer, you can go with a single selectMember function.

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;
}