Question

template <class T>
class Foo {
public:
    T val;
    Foo(T _v): val(_v){}
    friend ostream& operator<< (ostream& out, const Foo& A) {
        out << A.val;
        return out;
    }
};

template <class X, class Y> Foo<???> operator+(const Foo<X>& A, const Foo<Y> & B) {
    if (sizeof(X) > sizeof (Y))
        return Foo<X>(A.val + B.val);
    else
        return Foo<Y>(A.val + B.val);
}

int main() {
    Foo<double> a(1.5);
    Foo<int> b(2);
    cout << a << endl;
    cout << b << endl;
    cout << a+b << endl;
}

我的目标是让operator+函数根据参数的类型返回不同的类型。

例如，如果a为int且b为int，则返回Foo<int>，如果其中一个或两者为double，则返回{{ 1}}。

有可能吗？

Answer 1

（C ++ 11）：在dectype中使用declval表达式：

#include <utility>

template <class X, class Y> 
Foo<decltype(std::declval<X>() + std::declval<Y>())> operator+(...);

Answer 2

这是可能的（在C ++ 03或C ++ 11中）使用部分模板特化。

// C++ does not allow partial specialization of function templates,
// so we're using a class template here.

template <typename X, typename Y, bool xLarger>
struct DoPlusImpl // When x is selected
{
    typedef Foo<X> result_type;
};

template <typename X, typename Y>
struct DoPlusImpl<X, Y, false> // When y is selected
{
    typedef Foo<Y> result_type;
};

template <typename X, typename Y> // Select X or Y based on their size.
struct DoPlus : public DoPlusImpl<X, Y, (sizeof (X) > sizeof (Y))>
{};

// Use template metafunction "DoPlus" to figure out what the type should be.
// (Note no runtime check of sizes, even in nonoptimized builds!)
template <class X, class Y>
typename DoPlus<X, Y>::result_type operator+(const Foo<X>& A, const Foo<Y> & B) {
     return typename DoPlus<X, Y>::result_type
         (A.val + B.val);
}

您可以在IDEOne上看到这一点 - ＆gt; http://ideone.com/5YE3dg

Answer 3

是的！如果你想给出自己的规则，这就是你的方法：

template <typename X, typename Y> struct Rule {};
template<> struct Rule<int, int> { typedef int type;};
template<> struct Rule<float, int> { typedef bool type;};

然后

template <class X, class Y> Foo<typename Rule<X, Y>::type> operator+(...)

C ++模板自动类型推广

3 个答案: