c – 根据模板参数选择联合成员

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 <typ@R_944_8371@ 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 <typ@R_944_8371@ T>
T& SELEctMember(Union& u)
{
    return *((T*)&u);
}

template <typ@R_944_8371@ 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;
}